A molecule that contains three identical polar bonds to the central atom will be?
The question is incomplete, the complete question is;
A molecule that contains three identical polar bonds to the central atom will be ________.
nonpolar if the geometry is planar triangular
polar in all cases
nonpolar in all cases
impossible to tell the polarity
either polar or nonpolar depending on the identity of the atoms bonded to the central atom
Answer:
Nonpolar if the geometry is planar triangular
Explanation:
The polarity of molecules depends both on the polarity of individual bonds in the molecule as well as the overall dipole moment of the molecule. We must remember that dipole moment is a vector quantity hence direction of the resultant vector is very important.
Now, if i have a molecule that contains three identical polar bonds, a planar triangular geometry means that the molecule is symmetrical and will have an overall dipole moment of zero. Hence the molecule is nonpolar.
Help help help help help
How many Joules of heat are required to raise the temperature of 20.0 grams of water from 30.0oC to 40.0oC?
Answer:
840 J
Explanation:
c ≈ 4200 J / (kg * °C)
m = 20 g = 0,02 kg
[tex]t_{1}[/tex] = 30 °C
[tex]t_{2}[/tex] = 40 °C
The formula is: Q = c * m * ([tex]t_{2} - t_{1}[/tex])
Calculating:
Q = 4200 * 0,02 * (40 - 30) = 840 (J)
What is the mass in grams of 1.00 x 10 24 atoms of Mn?
a)91.3 g
b) 123.4 g
c) 1.66 g
d) 166 g
91.2 g Mn
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Reading a Periodic TableAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
[Given] 1.00 × 10²⁴ atoms Mn
Step 2: Identify Conversions
Avogadro's Numer
[PT] Molar Mass of Mn - 54.94 g/mol
Step 3: Convert
[DA] Set up: [tex]\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})[/tex][DA] Multiply/Divide [Cancel out units]: [tex]\displaystyle 91.2321 \ g \ Mn[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
91.2321 g Mn ≈ 91.2 g Mn
Gold has a density of 19.3 g/cm'. Find the volume, in cubic
centimeters, of a sample of gold that weighs 0.715 kg.
Answer:
37.05 cm³
Explanation:
0.715 kg = (0.715 × 1000) g = 715g
volume = mass/density = 715/19.3 = 37.05 cm³
Both diamond and graphite (i.e. pencil lead) consist of carbon atoms. They are only different in their crystalline structures. One carat or 0.20 g, of a high-quality diamond costs up to $5000, while 25 g of pencil lead may only cost $2. Determine the numbers of C-atom in diamond vs. graphite which can be obtained with $1.
Answer:
Moles of carbon atoms = 3.33 × [tex]10^{-6}[/tex] mol
No. of atoms of C in Diamond = 2.007 × [tex]10^{28}[/tex] atom
Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms
Explanation:
given data
Cost of 0.2g of diamond = $5000
Cost of 25 g of graphite = $ 2
solution
we know cost of 0.2g of diamond is $ 5000 so that for 1$
if buy 1$ = [tex]\frac{0.20}{5000}[/tex]
1$ = 4.0 × [tex]10^{-5}[/tex] g Carbon
and Moles of carbon atoms is express as
Moles of carbon atoms = Given mass of Carbon ÷ atomic mass of C .........1
Moles of carbon atoms = 4.0 × [tex]10^{-5}[/tex] g/ 2.0g
Moles of carbon atoms = 3.33 × [tex]10^{-6}[/tex] mol
and
No. of atoms of C in Diamond = No. of moles × Avogadro NO ..............2
No. of atoms of C in Diamond = 3.33 × [tex]10^{-6}[/tex] mol × 6.022 × [tex]10^{28}[/tex]
No. of atoms of C in Diamond = 2.007 × [tex]10^{28}[/tex] atom
Graphite
and wew have given Cost of 25 g of graphite is $2 so for but 1$ we get
for buy $1 = 25÷2 = 12.5 g Of graphite
Moles of graphite = 12.5÷12 = 1.04 mol
Atoms of graphite = 1.04 × 6.022 × 1023
Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms
What is the boiling point of a solution formed by dissolving 0.75 mol of KCl in 1.00 kg of water?
The boiling point of water generally increases as the amount of impurities (which a solute like KCl technically can be thought of) dissolved increases. This relation can be quantified using the equation,
[tex]\Delta T_b = i \times K_b \times m[/tex]
where [tex]\Delta{T}_{b}[/tex] is the change in the water's boiling point (normally taken to be 100 °C), [tex]i[/tex] is the Van 't Hoff factor (the number of particles a single formula unit of the solute dissociates into in water), [tex]K_b[/tex] is the boiling point elevation constant, and [tex]m[/tex] is the molality (moles of solute/kilogram(s) of solvent) of the solution.
We are forming a solution by dissolving KCl in water. KCl is an electrolyte that, in water, will dissociate into K⁺ and Cl⁻ ions. So, for every formula unit, KCl, we obtain two particles. Thus, the Van 't Hoff factor, or [tex]i[/tex], will be 2.
The molality of the solution can be calculated by dividing the number of moles of KCl by the mass of water in kilograms. Since we have 1.00 kg of water, we would be dividing 0.75 mol KCl by 1, giving us a molality (m) of 0.75 m.
We aren't provided the boiling point elevation constant for water. Several authoritative sources give the value 0.512 °C/m, so we will adopt that as our [tex]K_b[/tex].
Note: m = mol/kg as used in this problem.
Plugging everything in,
[tex]\Delta T_b = i \times K_b \times m \\\Delta T_b = 2 \times 0.512 \text{ } \frac{^oC}{mol/kg} \times 0.75 \text{ } \frac{mol}{kg} \\\Delta T_b = 0.768 \text{ } \mathrm{ ^oC}[/tex]
As you can see, our change in boiling point is positive (the boiling point is elevated), and it is also quite modest. Taking 100 °C to be the boiling point of pure water, the boiling point of our solution would be 100 ⁰C + 0.768 ⁰C, or 100.768 ⁰C.
If we are considering significant figures, then we must give our answer to two significant figures (since 0.75 has two sig figs). We can regard the boiling point of water (100 ⁰C) as a defined value. Since our final answer is a sum, the boiling point of our solution to two significant figures would be 100.77 ⁰C.
Given:
Mol = 0.75Mass = 1.00 kgWe know,
Boiling point constant, Kb = 0.51The molality of the solution will be:
= [tex]\frac{Mole}{Mass}[/tex]
= [tex]\frac{0.75}{1}[/tex]
= [tex]0.75 \ m[/tex]
Now,
→ [tex]T_{solution}-T_{water} = Kb\times m\times i[/tex]
By putting the values, we get
[tex]= 0.51\times 0.75\times 2[/tex]
[tex]= 0.765[/tex]
Boiling point of water = 100°Chence,
Solution's boiling point will be:
→ [tex]T_{solution} = 100+0.765[/tex]
[tex]= 100.765^{\circ} C[/tex]
Thus the above approach is right.
Learn more about boiling point here:
https://brainly.com/question/23549697
Which of the following would be produced when two or more different atoms bond chemically?
O A. compounds
.
B. elements
C. solutions
D. mixtures
Answer:
When two or more atoms chemically bond with each other, the resultant chemical structure is a molecule. The familiar water molecule, H2O, consists of two hydrogen atoms and one oxygen atom; these bond together to form water, irdk but it's either C or D
Explanation:
hope this helps have a good rest of your day :) ❤
A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the same amount of air is placed in a box with a volume of 5.0m3 at 35∘C? Report your answer with two significant figures.
Answer:
Given:
Initial pressure: [tex]3.65\; \rm atm[/tex].Volume was reduced from [tex]45\; \rm m^{3}[/tex] to [tex]5.0\; \rm m^{3}[/tex].Temperature was raised from [tex]25\; ^\circ \rm C[/tex] to [tex]35\; ^\circ \rm C[/tex].New pressure: approximately [tex]3.4\times 10\; \rm atm[/tex] ([tex]34\; \rm atm[/tex].) (Assuming that the gas is an ideal gas.)
Explanation:
Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:
Reduce the volume of the gas from [tex]45\; \rm m^{3}[/tex] to [tex]5.0\; \rm m^{3}[/tex]. Calculate the new pressure, [tex]P_1[/tex].Raise the temperature of the gas from [tex]25\; ^\circ \rm C[/tex] to [tex]35\; ^\circ \rm C[/tex]. Calculate the final pressure, [tex]P_2[/tex].By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)
For this gas, [tex]V_0 = 45\; \rm m^{3}[/tex] while [tex]V_1 = 5.0\; \rm m^{3}[/tex].
Let [tex]P_0[/tex] denote the pressure of this gas before the volume change ([tex]P_0 = 3.65\; \rm atm[/tex].) Let [tex]P_1[/tex] denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between [tex]P_1\![/tex] and [tex]P_0\![/tex]:
[tex]\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0[/tex].
In other words, because the final volume is [tex](1/9)[/tex] of the initial volume, the final pressure is [tex]9[/tex] times the initial pressure. Therefore:
[tex]\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm[/tex].
On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)
Convert the unit of the temperature of this gas to degrees Kelvins:
[tex]T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K[/tex].
[tex]T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K[/tex].
Let [tex]P_1[/tex] denote the pressure of this gas before this temperature change ([tex]P_1 = 32.85\; \rm atm[/tex].) Let [tex]P_2[/tex] denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at [tex]V_2 = V_1 = 5.0\; \rm m^{3}[/tex].
Apply Amonton's Law to find the ratio between [tex]P_2[/tex] and [tex]P_1[/tex]:
[tex]\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}[/tex].
Calculate [tex]P_2[/tex], the final pressure of this gas:
[tex]\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}[/tex].
In other words, the pressure of this gas after the volume and the temperature changes would be approximately [tex]3.4\times 10\; \rm atm[/tex].
According to the Bohr model of the atom, which particles are allowed to exist in any one of a number of energy levels?
Answer: the line-emission spectrum of an atom is caused by the energies released when electrons. releases energy of only certain values.
Convert 3.01 x 10^24 molecules of ammonium sulfate to mass
Mass of ammonium sulfate = 660.7 g
Further explanationGiven
3.01 x 10²⁴ molecules of ammonium sulfate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol ammonium sulfate (NH₄)₂SO₄ :
n = N : No
n = 3.01 x 10²⁴ : 6.02 x 10²³
n = 5
mass ammonium sulfate :
= mol x MW
= 5 x 132,14 g/mol
= 660.7 g
How many molecules are there in 45.0 grams of CH ?
I
How many molecules are there in 150.0 grams of CzHs?
Answer:
There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry
Explanation:
A chemistry student is given 2.00 L of a clear aqueous solution at 43.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 25.° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.062 kg1) Using only the information above, can you calculate the solubility of X in water at 25 degrees C?2) If yes calculate it. Round answer to 2 significant digits
Answer:
Follows are the solution to the given points:
Explanation:
In part 1:
As described and in the query, they become precipitated whenever the solutions are refrozen to [tex]25^{\circ} \ C[/tex].
Afterward, certain precipitate becomes replaced as well as the remaining water is evaporated, it implies that certain precipitate remained throughout the solution to just the container when the entire balance is evaporated.
The unrecoverable salt precipitates whenever the solvent is cooled at [tex]25^{\circ} \ C[/tex]and the remaining salt dissolves. It dissolved salt remains whenever the water is evaporated because as dissolved salt value is given that results can be achieved.
In part 2:
They have precipitation weight = [tex]0.063\ g[/tex]. They have a [tex]2 \ L[/tex] the solution, they may disregard the volume increases due to its precipitation. The intensity therefore is [tex]\frac{0.063}{2} = 0.0315 \ \frac{g}{L}[/tex]
HELP ME PLS I WILL GIVE BRAINLYEST 1. Is this organism affecting the lives of humans? 2. How is it affecting the lives of humans? 3. What are some ways to prevent the spread of zebra mussels?
Answer:
Yes it is because it effects the food chain in many ways like it takes more time to get rid of them and they eat parasites that other living things need. We can get rid of zebra mussels by removing them little by little or putting animals in the water that eats them.
Explanation:
Answer:
By encouraging boaters to carefully clen, drain and dry their boats before launching them in different bodies of water.
Explanation:
Draw a mechanism for the reaction of methanol with sodium amide. In the box to the left, draw any necessary curved arrows. Show the products of the reaction in the box to the right. Include any nonzero formal charges and all lone pairs of electrons. Finally, check the box to indicate which side of the reaction is favored at equilibrium.
Answer:
See explanation
Explanation:
When sodamide reacts with methanol, the following equilibrium is set up;
CH3OH(aq) + NH2-(aq) ⇄ CH3O-(aq) + NH3(aq)
In the reaction, the sodamide acts as a base that abstracts a proton from methanol to yield the conjugate base of methanol which is the methoxide ion.
The forward reaction is favoured at equilibrium.
2. Describe briefly of some ways these organisms might interact with
each other. (5 points)
I need help with this!!!
Answer:
0.73g/cm^3
Explanation:
d=m/v
d=11/15
d=0.73
Calculate the [H+]
and pH of a 0.000295 M
butanoic acid solution. Keep in mind that the a
of butanoic acid is 1.52×10−5
[H⁺]=6.696 x 10⁻⁵
pH = 4.174
Further explanationGiven
The concentration of 0.000295 M (2.95 x 10⁻⁴ M) butanoic acid solution
Required
the [H+] and pH
Solution
Butanoic acid is the carboxylic acid group. Carboxylic acids are weak acids
For weak acid :
[tex]\tt [H^+]=\sqrt{Ka.M}[/tex]
Input the value :
[H⁺]=√1.52 x 10⁻⁵ x 2.95 x 10⁻⁴
[H⁺]=6.696 x 10⁻⁵
pH = - log [H⁺]
pH = - log 6.696 x 10⁻⁵
pH = 5 - log 6.696
pH = 4.174
Start with 3 breads and 2 cheeses. How many products are made? 1 What are the leftovers?
Answer:
1 product is made. The left overs are 1 piece of bread and 1 slice of cheese
Explanation:
what other traits besides phisical ones could be passed on from parent offspring
Answer:
Love for Music
Explanation:
This is one example of many non-physical traits. In the womb a mother can listen to her favorite music and the growing baby could grow to like it in the womb!
This is just one of the many other traits that could be passed to their offspring.
Hope this Helps!
ch4(g) + h2o(g) 3h2(g) + co(g) enthalpy of formation of CH4
Answer:
Kc=[[CO][H2]3[CH4][H2O]
3.90=(0.30)(0.10)3[CH4]×0.02
[CH4]=0.023.90×0.30×(0.10)3=5.85×10−2 M
Thus, the concentration of methane in the mixture is 5.85×10−2 M.
Please answer, this is due in 30 minutes
Answer:
0.591 g of magnesium phosphate is the theoretical yield.
Magnesium nitrate is the limiting reactant.
Explanation:
Hello!
In this case, since the balanced reaction turns out:
[tex]3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3[/tex]
Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:
[tex]m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2[/tex]
Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.
However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:
[tex]Y=\frac{actual}{0.591g}*100\%[/tex]
Best regards!
For each bond, show the direction of polarity by placing a + sign next to the atom expected to have a partial positive charge and a − sign next to the atom expected to have a partial negative charge.
(A) { } Ge-Se { }
(B) { } Ge-Br { }
(C) { } Br-Se { }
Answer:
A) { + } Ge-Se { -}
B) { + } Ge-Br { - }
C) { - } Br-Se { + }
Explanation:
The (-)ive sign shall be placed for the atom with higher electronegativity, while the other atom will be electropositive.
a) Electronegativity of Ge = 2.01
Electronegativity of Se = 2.55
{ + } Ge-Se { -}
b) Electronegativity of Ge = 2.01
Electronegativity of Br = 2.95
{ + } Ge-Br { - }
c) Electronegativity of Br = 2.95
Electronegativity of Se = 2.55
{ - } Br-Se { + }
If energy is conserved will the maximum speed of the pendulum depend on the mass, the length, or both? How?
Answer:
BUDDY YO
Explanation:
please help me with this problem
the amount of reactants = the amount of products
N2 + 3H2 --> 2NH3
Can someone help me with this
Answer:
wow!
5. C
6. B
7. B
8. A
Explanation:
General equilibrium problems. ICE type problems.a. Isopropyl alcohol can dissociate into acetone and hydrogen according to the reaction below.At 179 °C, the equilibrium constant for this dehydrogenation reaction is 0.444. i) If 0.166moles of isopropyl alcohol is placed in a 10 L vessel and heated to 179 °C, what is the partialpressure of acetone when equilibrium is attained
Answer:
Explanation:
In a gaseous reaction mixture partial pressure is proportion to mole of the gas concerned .
Pressure of the reactant gas from gas equation
PV = nRT
P = nRT / V
= .166 x .082 x ( 273+179) / 10
= .615 atm
C₃H₇OH = (CH₃)₂CO + H₂
before reaction moles in terms of pressure
.615 0 0
After reaction
.615 - x x x
.444 = x² / ( .615 - x )
.273 - .444 x = x²
x² + .444 x - .273 = 0
x = .361 atm
So partial pressure of acetone is .361 atm at equilibrium.
how does a constructive relationship influence your well being
Answer:
how does a constructive relationship influence your well being
Explanation:
Constructive and deconstructive relationships can influence the well being of different people in different ways. From the point of view who value relationships, dealing with one who repeatedly ruins it is stressful, it slowly damages the health and will continue to do so unless the individual separates himself or herself from the one causing the pain.
Potassium nitrate (KNO3) is a water-soluble white powder that
is often used as a plant fertilizer. What is the molar
concentration of a solution made up of 505 grams of
potassium nitrate mixed with 250 mL water? The molar mass
of KNO3 is 101.1 g/mol. Round your answer to 3 sig figs.
The molar concentration, often called molarity, describes how much of a substance (a solute) is present per unit of solvent. By definition, the molar concentration (M) is equal to the number of moles (n) of solute divided by the number of liters (the volume, or V) of the solution.
Here, your solute is potassium nitrate, or KNO3. You're given the mass of KNO3 (505 g), but you need to convert this quantity to moles before you can find the molarity. To go from mass to moles, simply divide the mass of the substance by its molar mass (given to you as 101.1 g/mol).
[tex]505 \text{ g KNO}_\text{3} \div 101.1 \text{ g KNO}_3/\text{mol KNO}_{3} = 4.995 \text{ mol KNO}_3[/tex]
Now that we have the moles of solute, we divide by the liters of solution. We're given the volume of solution in milliliters, so to convert to liters, simply divide by 1000 (1 L = 1000 mL, so 1 mL = 1/1000 mL). Our volume of solution is thus 0.250 L.
Finally, we can calculate the molar concentration of the KNO3 solution:
[tex]4.995 \text{ mol KNO}_{3 }\div 0.250 \text{ L} = 19.98 \text{ }\frac{\text{moles}}{\text{liters}}\text{ of KNO}_3[/tex]
But, we're told to round our answer to three sig figs. Thus, our rounded and final answer would be 20.0 moles/liters of KNO3.
How did American prepare for and fight for the France battle
Answer:
their warfare had changed dramatically
Answer:
Explanation:
They first tried to build there army. They did this by passing the Selective Service Act. ... In the next 18 months about 4 million men and women joined the army, but then the army had to be educated.