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How many moles of Ag+ ions are present in 10 ml of a 0.75 M AgNO3 solution?

lana66690 [7]

Answer:

b. 7.5 x 10^-3

Explanation:

To solve this problem we need to keep in mind the definition of molarity:

Molarity = moles of solute / liters of solution

With the above information in mind it is possible to calculate the moles of solute, given the volume (10 mL) and concentration (0.75 M) of the solution:

First we convert 10 mL to L ⇒ 10 mL / 1000 = 0.01 L

Then we calculate the moles of AgNO₃:

moles of solute = Molarity * Liters of solution

0.01 L * 0.75 M = 7.5x10⁻³ mol AgNO₃

One mole of AgNO₃ contains one mole of Ag⁺, thus the number of Ag⁺ moles is also 7.5x10⁻³.

5 0

3 years ago

Free brainiest! the first person who makes me blush or makes me want to kiss you.

lawyer [7]

Answer:

Explanation:Do you like to draw coz you put the D in raw

3 0

3 years ago

Read 2 more answers

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the den

NISA [10]

Answer:

$\rho=1.54\ g/cm^3$

Explanation:

The expression for density is:

$\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}$

$N_a=6.023\times 10^{23}\ {mol}^{-1}$

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

$\rho$ is the density

Radius = 197 pm = $1.97\times 10^{-8}\ cm$

Also, for fcc, $Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm$

Thus,

$\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3$

$\rho=\frac{160.312}{104.19483}\ g/cm^3$

$\rho=1.54\ g/cm^3$

7 0

3 years ago

Can someone answer 57 and 58

RSB [31]

Any type of electromagnetic energy's wavelengths that are shorter than a type of electromagnetic energy's wavelengths is to the left of that type of electromagnetic energy on the diagram. Hence, the wavelength of a type of energy decreases as you progress to the left on the diagram and left of the green light (in the visible spectrum).

4 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: The percent yield of water in the reaction is 46.85 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaOH:

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol$

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = $\frac{2}{1}\times 0.741=1.482moles$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%$

Hence, the percent yield of water in the reaction is 46.85 %.

8 0

3 years ago