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2 years ago

What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)

Chemistry

1 answer:

seraphim [82]2 years ago

6 0

Answer:

219.15 grams

Explanation:

What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)

Mole of Na = 22.99g

Mole of Cl = 35.45g

For NaCl we have ratio of 1:1, so we have 1 Na for every Cl

So we just add the two together to get the molar mass of NaCl which is

22.99 + 35.45 = 58.44g/mol

And we know we have 3.75 moles of NaCl so we multiply that by the molar mass of NaCl to get our answer

3.75 x 58.44 = 219.15grams

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8 months ago

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

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

<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

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2 years ago

As changes in energy levels of electrons increase, the frequencies of atomic line spectra they emit ___?

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Ans: As changes in energy levels of electrons increase, the frequencies of atomic line spectra they emit will <u>increase.</u>

The energy (E) is related to the frequency (ν) by the following equation:

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Based on the above equation, as the changes in energy levels increase, the frequency of emitted radiation will also increase.

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3 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago