Answer:
true once light has been produce it will keep travelling jn straight parts until it hits something else
separete the ALKALI from the following bases :
NH4OH(ammonium nitrate)
CuO(copper oxide)
Zn(OH)2 (zinc hydroxide)
MgO(magnesium oxide)
Na2O(sodium oxide)
NaOH(sodium hydroxide)
CoO(cobalt oxide)
Mg(OH)2(magnesium hydroxide)
LIOH(lithium hydroxide)
help me with this i will surely mark u as Brainliest
plss help!!!
Answer:
Ammonium hydroxide, NH₄OH
Magnesium hydroxide, Mg(OH)₂
Sodium hydroxide, NaOH
Lithium hydroxide, LiOH
Explanation:
A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.
Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.
Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.
From the given chemical compounds the alkalis present in the list are:
Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions
Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions
Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions
Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions
CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.
Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble
MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.
Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.
CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.
Select the net ionic equation for the reaction that occurs when magnesium sulfate and nickel(II) nitrate are mixed.
a. Ni2+(aq) + SO4^2- → NISO2 (s) + O2 (g).
b. Mg2+(aq) + 2NO3 (aq) → Mg(NO3)2(s).
c. Mg2+(aq) + NO3- → MgNO3 (s).
d. Mg2+(aq) + SO4^2- (aq) + Ni2+ (aq) + 2NO3- → Mg2+ (aq) + 2NO3 (aq) + NISO4 (s).
e. Ni2+(aq) + SO4^2- (aq) → NISO4 (s).
f. No reaction occurs.
Answer:
No reaction occurs.
Explanation:
The molecular reaction is as follows;
MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)
We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.
This reaction does not occur since the two products that ought to be obtained are soluble in water.
Soda contains phosphoric acid (H3PO4). To determine the concentration of phosphoric acid in 50.0 mL of soda, the available phosphate ions are precipitated with excess silver nitrate as silver phosphate (418.58 g/mol). The dry Ag3PO4 is found to have a mass of 0.0576 g. What is the concentration of phosphoric acid in the soda?
Answer:
0.0270w/v% H3PO4 in the soda
Explanation:
All phosphates reacts producing Ag3PO4. To solve this question we must convert the mass of Ag3PO4 to moles. These moles = moles of H3PO4. We can find, thus, the mass of H3PO4 and the w/v% as follows:
Moles Ag3PO4 -Molar mass: 418.58g/mol-
0.0576g * (1mol / 418.58g) = 1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4
Mass H3PO4 -Molar mass: 97.994g/mol-
1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4 * (97.994g/mol) = 0.0135g H3PO4
w/v%:
0.0135g H3PO4 / 50.0mL * 100 =
0.0270w/v% H3PO4 in the sodaAll the properties listed below are characteristic of the transition elements except __. a) most are paramagnetic b) most are colored c) most have high electronegativities d) most have multiple oxidation states e) most form many different complexes
Answer:
c) most have high electronegativities
Explanation:
They tend to have high electric CONDUCTIVITY because of the free-flowing d-orbital electrons, but have low electron affinity, ionization energy, and electronegativities.
Sợ sánh kỹ thuật trích ly rắn lỏng với kỹ thuật trích ly lỏng
lỏng
Answer:
which language is this please tell me
mẫu khi thêm NH4NO3 vào đem nung để nguội lại thêm NH4NO3 có tác dụng gì?
Answer:
Adding ammonium nitrate to water turns the mixture cold and is a good example of an endothermic chemical reaction!
Spell out the full name of the compounds
Help plz
Answer:
propanal
Explanation:
hope this helps :)
Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO
Answer:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
3 NO₂ + H₂O → 2 HNO₃ + NO
Explanation:
We will balance the equation using the trial and error method.
C₂H₅OH + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.
C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O
2) We balance O atoms by multiplying O₂ by 3.
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
NH₃ + O₂ → NO₂ + H₂O
1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.
2 NH₃ + O₂ → NO₂ + 3 H₂O
2) We balance N atoms by multiplying NO₂ by 2.
2 NH₃ + O₂ → 2 NO₂ + 3 H₂O
3) We balance O atoms by multiplying O₂ by 3.5
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
2) We balance O atoms by multiplying O₂ by 5.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + NaOH → Na₂SO₄ + H₂O
1) We balance Na atoms by multiplying NaOH by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O
2) We balance H and O atoms by multiplying H₂O by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
NO₂ + H₂O → HNO₃ + NO
1) We balance H atoms by multiplying HNO₃ by 2.
NO₂ + H₂O → 2 HNO₃ + NO
2) We balance N atoms by multiplying NO₂ by 3.
3 NO₂ + H₂O → 2 HNO₃ + NO
A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/goC.)
Answer: The final temperature will be [tex]52.74^oC[/tex]
Explanation:
Calculating the heat released or absorbed for the process:
[tex]q=m\times C\times (T_2-T_1)[/tex]
In a system, the total amount of heat released is equal to the total amount of heat absorbed.
[tex]q_1=-q_2[/tex]
OR
[tex]m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)[/tex] ......(1)
where,
C = heat capacity of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of water of sample 1 = 100.0 g
[tex]m_2[/tex] = mass of water of sample 2 = 71.0 g
[tex]T_f[/tex] = final temperature of the system = ?
[tex]T_1[/tex] = initial temperature of water of sample 1 = [tex]27^oC[/tex]
[tex]T_2[/tex] = initial temperature of the water of sample 2 = [tex]89.0^oC[/tex]
Putting values in equation 1, we get:
[tex]100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC[/tex]
Hence, the final temperature will be [tex]52.74^oC[/tex]
____________ can increase the presence of 5HT in the terminal button or synaptic cleft.
Answer:
Selective serotonin reuptake inhibitors (SSRIs)
Explanation:
A synaptic cleft is a space that separates two neurons thereby forming a junction between two or more neurons. The synaptic cleft helps in the transfer of nerve impulse from one neuron to the other.
5-HT is found in the enteric nervous system located in the gastrointestinal tract and it helps in modulating cognition, memory, sleep, and numerous physiological processes.
Selective serotonin reuptake inhibitors (SSRIs), such as fluoxetine and citalopram are used to increase the level of 5-HT in the synaptic cleft by inhibiting its reuptake into the presynaptic terminal.
You have run an experiment studying the effects of the molecular weight of a compound on the rate of diffusion in agar. Compound X has a molecular weight of 25.3 g/mol and compound Y has a molecular mass of 156.2 g/mol. On two separate agar plates, 0.1 g of each substance were transferred and allowed to diffuse for 2 hours. The results below were obtained. Use this information about this situation to answer the following three questions.
a. What was the independent variable in this experiment?
b. What was the independent variable in this experiment?
c. List two controls that were held constant for this experiment or that you would hold constant for this experiment
Answer:
A) Molecular weight
B) Rate of diffusion
C) Agar plates
Time of diffusion
Explanation:
A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.
B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.
C) A control variable is one that would be held constant throughout the research.
In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.
The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) list carbon only Chirulit,
Answer:
(d)
Explanation:
Carbonyl group can be the placement of kerosene sugar
When zinc nitrate is heated, zinc oxide, nitrogen dioxide(NO2) and oxygen gas are
produced.
i.
Calculate the mass of Zinc oxide produced if 38.5 g of zinc nitrate is heated.
ii.
Determine the volume of Nitrogen dioxide gas evolved at rtp
Answer:
i) 16.5g of ZnO
ii) 9.8 dm³ of NO2
Explanation:
The working is shown in the photo so kindly refer to it
Which best describes how the total mass of the substances that go into
photosynthesis compares to the mass of substances that are present
afterward?
O A. The mass increases because the molecules that are produced are
larger than those that are used.
B. The mass increases because some light energy changes into
mass.
O C. The mass stays the same because the total number of atoms
does not change
O D. The mass decreases because plants destroy some of the atoms
during photosynthesis.
Answer:
C. The mass stays the same because the total number of atoms does not change
Explanation:
According to the law of conservation of matter/mass, matter cannot be created nor destroyed, hence, the amount of matter in the reactants must be the same amount in the products.
Using the photosynthetic reaction as a case study, carbon dioxide (CO2) and water (H2O) are the compounds that go into the reaction (reactants) while glucose and oxygen (O2) are the products of the reaction.
Using the law of conservation of matter to explain, the total mass of both the reactants and products stays the same because the total number of atoms does not change i.e. if 6 atoms of Carbon starts the reaction, 6 atoms of carbon will end it.
Suppose you are trying to separate three Proteins using Gel-Filtration chromatography. The sizes of each are given below:
Protein A: 1200 kDa
Protein B: 2000 kDa
Protein C: 800 kDa
Which protein will be the first to emerge from the column?
Answer: The correct answer is Protein B.
Explanation:
Gel-filtration chromatography is a separation technique that is based on the size of the molecules in a compound. It is also known as size-exclusion chromatography in which the eluent (carrier) used is an aqueous solution.
The matrix that is used is a porous material. When the sample is inserted in the column, the smaller particles interact strongly with the matrix than the large ones. Thus, as the eluent is passed through the matrix, larger molecules come out first, and the smallest molecule comes out last.
Given sizes of the proteins:
Protein A: 1200 kDa
Protein B: 2000 kDa
Protein C: 800 kDa
As protein B has the largest size of all the given proteins, it will emerge out first from the column.
Hence, the correct answer is Protein B.
For a particular chemical reaction the rate (g/hr) at which one of the reactants changes is proportional to the amount of that reactant present. If y represents the amount of that reactant at time t, StartFraction dy Over dt EndFraction equals minus0.7y. If there were 70 grams of the reactant when the process started (tequals 0), how many grams will remain after 4 hours?
Answer:
Amount of reactant after four hours = 4,26 grams
Explanation:
Suppose y denotes the amount of reactant at the time (t)
The given function:
[tex]\dfrac{dy}{dt} = -0.7 y[/tex]
[tex]\dfrac{dy}{y} = -0.7 dt[/tex]
Taking integral on both sides
㏑(y) = -0.7t + c¹
[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]
[tex]y(t) = Ce ^{-0.7t}[/tex]
At t = 0 ; y (t) = 70
∴
[tex]70 = Ce^{-0.7(0)}[/tex]
C = 70
As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]
After four hours, the amount of the reactant is:
[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]
[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]
[tex]\mathtt{y(t) = 4.26}[/tex]
Amount of reactant after four hours = 4,26 grams
Solution A has a pH of 7, and solution B has a pH of 14. Which statement
best describes these solutions?
Answer:
Option A. Solution B is basic, and solution A is neutral.
Explanation:
The pH of a solution is simply defined as the measure of acidity or alkalinity of a solution.
The pH scale ranges from 0 to 14 with the following readings:
0 to 6 => Acidic solution
7 => Neutral solution
8 to 14 => Alkaline / basic solution.
From the above, we understood that solutions with pH ranging from 0 to 6 are acidic solutions. Those with pH of 7 are neutral solutions while those with pH ranging from 8 to 14 are basic solutions.
With the above information in mind, let us answer the question given above. This is illustrated below:
pH of solution A = 7
pH of solution B = 14
Solution A has a pH of 7. This implies that solution A is a neutral solution
Solution B has a pH of 14. This implies that solution B is a basic solution.
Thus, option A gives the correct answer to the question.
What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?
Answer:
37.9% v/v
Explanation:
Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):
volume concentration = volume of solute / volume of solution
[tex]\% v/v = 55/145= 0.379[/tex]
The rate of the reaction is 1.6*10-2 M/s when the concentration of A is 0.15 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.
Answer:
[tex]k_1=0.107s^{-1} \\\\k_2=0.711M^{-1}s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information and the attached picture in which we can see the units of the rate constant, it turns out possible for us to realize the two called rate laws are:
[tex]r=k[A]\\\\r=k[A]^2[/tex]
The former is first-order and the latter second-order; in such a way, we solve for the rate constant in both cases to obtain the following:
[tex]k=\frac{r}{[A]}=\frac{1.6x10^{-2}M/s}{0.15M}=0.107s^{-1} \\\\k=\frac{r}{[A]^2}=\frac{1.6x10^{-2}M/s}{(0.15M)^2}=0.711M^{-1}s^{-1}[/tex]
Regards!
Describe why corrosion is a natural process
Answer :
Answer :because it happens due to moisture and oxygenFe có tác dụng với HCL không
Fe có tác dụng với Hcl
Fe + 2hcl -> fecl2 +h2
Answer:
có
Explanatio:
Classify each structure according to its functional class.
Compound A contains a carbonyl bonded to two alkyl groups.
Compound B contains an oxygen bonded to two alkyl groups.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound D is a nitrogen bonded to three alkyl groups.
Classify structure A according to its functional class.
Classify structure B according to its functional class.
Classify structure C according to its functional class.
Classify structure D according to its functional class.
Answer:
Classify each structure according to its functional class.
Compound A contains a carbonyl bonded to two alkyl groups.
Compound B contains an oxygen bonded to two alkyl groups.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound D is a nitrogen bonded to three alkyl groups.
Explanation:
Compound A contains a carbonyl bonded to two alkyl groups.
-C=O group is called a carbonyl group.
If it is present between two alkyl groups then, it is a ketone.
Compound B contains oxygen bonded to two alkyl groups.
Compound B is an example of an ether molecule.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound C is C3H7-CO-NHCH3 which is an amide molecule.
Compound D is nitrogen bonded to three alkyl groups.
This is an example of a tertiary amine group.
Boiling point-methanol (65.0) 66.8c.Boiling point-unknown (record from video)——-c
Identify of unknown:
Possibilities are:Mathanol65.0c;Ethanol 78.5c; Acetone 56.0C
Identify the conjugate acid/base pairs in each of the following equations:
(a) H2S + NH3 ⇔ NH4+ + HS-
Pair 1: H2S and
Pair 2: NH3 and
(b) HSO4- + NH3 ⇔ SO42- + NH4+
Pair 1: HSO4- and
Pair 2: NH3 and
(c) HBr + CH3O- ⇔ Br- + CH3OH
Pair 1: HBr and
Pair 2: CH3O- and
(d) HNO3 + H2O → NO3- + H3O+
Pair 1: HNO3 and
Pair 2: H2O and
Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
When stirred in 30°C water, 5 g of powdered potassium bromide, KBr, dissolves faster than 5 g of large crystals of potassium bromide. Which of the following best explains why the powdered KBr dissolves faster?
A. Potassium ions and bromide ions in the powder are smaller than potassium ions and bromide ions in the large crystals.
B. Powdered potassium bromide exposes more surface area to water molecules than large crystals of potassium bromide.
C. Fewer potassium ions and bromide ions have been separated from each other in the powder than in the crystals.
D. Powdered potassium bromide is less dense than large crystals of potassium bromide.
Answer:
B
Explanation:
Do diện tích tiếp xúc ở dạng bột cao hơn dạng tinh thể
What type of organic compound is a reactant in all substitution reactions?
1.alkyne
2.alkane
3.alkene
The organic compounds that are capable of being a reactant in all substitution reactions would belong to the alkane group.
Alkanes and substitution reactionsAlkanes are saturated compounds that ordinarily will not participate in addition reactions due to their saturated nature.
Thus, alkanes are only able to participate in substitution reactions involving the substitution of one or more of their component atoms for another atom.
This is unlike alkyne and alkenes which are naturally unsaturated. The unsaturation makes them a candidate for additional reactions.
More on saturation and substitution reactions can be found here: https://brainly.com/question/3735006
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Perform the following
mathematical operation, and
report the answer to the correct
number of significant figures.
7.125 x 8.00 = [?]
Answer ?
A company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield.
Both scientists started with 50.0 g of manganese oxide.
What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
Answer:
The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g
Explanation:
Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol
Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol
Moles of MnO₂ in 50 g = reacting mass / molar mass
where reacting mass = 50 g
Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1
Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄
Mass of 0.575 moles of KMnO₄ = number of moles × molar mass
Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄
Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g
He predicted an element with an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum that he named ekaboron. This element was discovered in 1879. What is this element:?
Answer:
Scandium
Explanation:
Mendeleev played an important role in the development of the modern periodic table. His periodic table was filled with gaps. He said that these gaps were elements that were yet to be discovered. He rightly predicted many elements which have now been discovered and fitted in their proper places in the periodic table.
He used the prefix ''eka'' to refer to elements whose properties were alike but were yet to be discovered at that time.
The compound named ekaboron which he predicted to have an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum was later discovered in 1879 and properly named scandium.
why Mg(OH)2 is soluble in HCL
Answer:
While Mg(OH)2 is practically insoluble, a certain amount of Mg(OH)2 dissociates into ions when put in water. ... As HCl is added to the beaker containing milk of magnesia, the H+ ions from the HCl react with the OH– ions (those that are actually in solution from the Mg(OH)2) according to Equation 3 below.