PLEASE ANSWER

4. True or false: You should use Avogadro's number to convert from moles to molecules

Dmitriy789 [7]

True, to convert from moles to atoms, multiply the molar amount by Avogadro's number. To convert from atoms to moles, divide the atom amount by Avogadro's number (or multiply by its reciprocal).

8 0

4 years ago

PLEASE HELP ME a) b) c) or d)?

PilotLPTM [1.2K]

Answer:

D / 15.0 g

Explanation:

3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions

.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle

5 0

3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

7 0

4 years ago

The reaction H2O+CH3Br=CH3OH+HBr. H2O acts as what

uysha [10]

Answer:

H2O acts as an oxidizing agent

3 0

2 years ago

How many liters of hydrogen gas is formed from the complete reaction of 15.2 g C? Assume that the hydrogen gas is collected at a

ira [324]

Answer:

38 L

Explanation:

There is some info missing. I think this is the original question.

Consider the chemical reaction: C(s) + H₂ O(g) ⟶ CO(g) + H₂ (g). How many liters of hydrogen gas is formed from the complete reaction of 15.2 g C? Assume that the hydrogen gas is collected at a pressure of 1.0 atm and a temperature of 360 K.

Step 1: Write the balanced equation

C(s) + H₂ O(g) ⟶ CO(g) + H₂ (g)

Step 2: Calculate the moles corresponding to 15.2 g of C

The molar mass of C is 12.01 g/mol.

$15.2g \times \frac{1mol}{12.01g} = 1.27 mol$

Step 3: Calculate the moles of H₂ produced from 1.27 moles of C

The molar ratio of H₂ to C is 1:1. The moles of H₂ produced are 1/1 × 1.27 mol = 1.27 mol.

Step 4: Calculate the volume of H₂

We will use the ideal gas equation.

$P \times V = n \times R \times T\\V = \frac{n \times R \times T}{P} = \frac{1.27mol \times \frac{0.0821atm.L}{mol.K} \times 360K}{1.0atm}= 38 L$

3 0

4 years ago