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Mekhanik [1.2K]
3 years ago
7

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how

much water should be mixed?
Chemistry
1 answer:
Aloiza [94]3 years ago
4 0

<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

<u>Explanation:</u>

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

\Rightarrow 16\times \frac{30}{100}=4.8L

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

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What is the mass of 8.23 x 10^23 atoms of Ag
Gnom [1K]

Answer:

\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}

8.23 *10^{23}  *\frac{1 \ mol \ Ag}{6.022*10^{23} }

Condense into one fraction.

\frac{8.23 *10^{23}  }{6.022*10^{23} } \ mol \ Ag

1.366655596 \ mol \ Ag

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

  • Ag: 107.868 g/mol

Let's make another ratio using this information.

\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

Multiply by the number of moles we calculated.

1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

The moles of silver cancel out.

1.366655596 *\frac {107.868 \ g \ Ag}{1 }

1.366655596 * {107.868 \ g \ Ag}

147.4184058 \ g\ Ag

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

  • 147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

147 \ g\ Ag

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0
3 years ago
Will marks brainly if it isn’t a copied answer!!!
Neporo4naja [7]

What kind of control like scientific?

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