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3 years ago

Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometry problem

1 answer:

3 0

Basically stoichiometry is the measurement of elements, that is, the study of chemical quantities consumed or produced in a chemical reaction. When we are performing it we are using a special chemical counting unit: the mole, a unit of measurement, and one mole of a substance contains 6.022 * 10^23 particles. Now mole ratio is defined as the ratio of moles of one substance to the moles of another substance in a balanced equation. If we are looking for the mole ratio between two substances, we need to look at the balanced equations for the coefficients in front of the substances you are interested in.
This should be your guiding mantra for doing stoichiometry problems!

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What is the final concentration of the solution produced when 225.5 mL of 0.09988-M solution of Na2CO3 is allowed to evaporate u

ikadub [295]

Explanation:

The given data is as follows.

$V_{1}$ = 225.5 mL, $V_{2}$ = 45.00 mL

$M_{1}$ = 0.09988 M, $M_{2}$ = ?

Therefore, formula to calculate final concentration will be as follows.

$M_{1}V_{1}$ = $M_{2}V_{2}$

Putting the given values into the above formula as follows.

$M_{1}V_{1}$ = $M_{2}V_{2}$

$0.09988 M \times 225.5 mL$ = $M_{2} \times 45.0 mL$

$M_{2}$ = 0.5 M

Thus, we can conclude that the final concentration of the given solution is 0.5 M.

4 0

3 years ago

Which statement best describes the effect of low ionization energies and low electronegativities on metallic bonding?

kramer

The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
Due to these low energies and negativities, valence electrons can be moved around quite easily and their positions may be altered quite drastically.

6 0

3 years ago

Read 2 more answers

Please help me for question 1 and 2

likoan [24]

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

4 0

3 years ago

Which chemical equations show a precipitation reaction?

Anestetic [448]

A and D

Hope this helps

6 0

3 years ago

Read 2 more answers

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago