Calculate the pressure exerted by 40 g of oxygen enclosed in a 1litre flask at 25°C. (0=16) (3 Marks) ​

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.

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.

Answer:

i think its c [everything is so blury]

Explanation:

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

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.

Answer:

electrophilic addition and

nucleophilic addition.

Answer:

the two types of addition compoundsare:

1.electrophilic addition

2. nucleophilic addition.

Answer:

which language is this please tell me

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.

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⁺.

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.

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]

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

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 =

Answer:

propanal

Explanation:

hope this helps :)

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

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

Answer:

Explanation:

This is because, Blue color highest frequency of energy after Violet and Indigo.

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.

Answer:

as in shapes

Explanation:

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.

Answer:

Adding ammonium nitrate to water turns the mixture cold and is a good example of an endothermic chemical reaction!

Fe có tác dụng với Hcl

Fe + 2hcl -> fecl2 +h2

Answer:

có

Explanatio:

Answer:

D

Explanation:

there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting

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!

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

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.

Answer:

The quantity of  heat required to raise the  temperature of 500 g of  iron by 2°C is 500 J.

Explanation:

Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state (solid, liquid or gaseous).

Its mathematical expression is the fundamental equation of calorimetry. This is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this  case:

Replacing:

Q= 500 [tex]\frac{J}{kg*C}[/tex] *0.500 kg*2 C

Solving:

Q= 500 J

The quantity of  heat required to raise the  temperature of 500 g of  iron by 2°C is 500 J.

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

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.

Answer:

20,8ml  NaCl 6M

Explanation:

C1V1 = C2V2