A solution is 12.5% Na3PO, by mass. How many grams of Na3PO4

Chemistry

1 answer:

Harman [31]3 years ago

7 0

<h2>Answer: 4.94 g </h2>

Explanation:

Mass of solution of Na₃PO₄ = volume × density

= 35.9 mL × 1.10 g/mL

= 39.49 g

Now, since the solution only contains 12.5% Na₃PO₄, then the mass of Na₃PO₄ is 12.5% of 39.49 g

12.5% × 39.49 g = <u>4.94 g </u>

You might be interested in

From what you've learned so far, how is molecular structure related to smell?

Sidana [21]

Answer:

Changing the shape of the molecules that create fragrances in a flower or fruit may influence our perception of their smell. The reaction pattern produced, olfactory code, is sent as a signal to the brain, which which is how you smell things.

Explanation:

7 0

3 years ago

How many moles of n are in 0.163 g of n2o?

zimovet [89]

First convert 0.163 grams of N2O to mol by dividing it with the molecular weight. The molecular weight of N2O is 44 grams/mol. The answer would be 3,79x10^-3. Then multiply it with 2 since there are 2 Nitrogen in one mole of N2O. Therefore, there are 7.41x10^-3 moles of Nitrogen.

3 0

3 years ago

Read 2 more answers

How many mols are present in the a sample of silver nitrate which has<br> 5.3x10^24 molecules.

Westkost [7]

Answer: 5.3 x 10^24 formula units of silver nitrate is equivalent to 8.8 moles of silver nitrate. Silver nitrate is an ionic compound, therefore, its representative particle is called a "formula unit" instead of molecule. For every mole of a substance, we know that there are 6.022 x 10^23 representative units of that substance. The amount of particles in one mole of substance is called Avogadro's number.

Further Explanation:

We can convert from number of representative particles to moles using the formula:

$\boxed {no. \ of \ moles \ = \ ( given \ no. \ of \ particles) \ (\frac{1 \ mole}{\ 6.022 \ x 10^{23} particles})}$

For this problem, we can calculate the number of moles by plugging in the given values to the equation above,

$no. \ of \ moles \ = (5.3 \ x \ 10^{24} \ formula \ units \ AgNO_{3}) \ (\frac{1 \ mole \ AgNO_{3}}{6.022 \ x 10^{23} \ formula \ units AgNO_{3}}) \\\\\boxed {no. \ of moles \ AgNO_{3} \ = \ 8.8 \ moles}$