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mezya [45]
3 years ago
6

If you have a density of 3g/ml and volume of 10mL.what is the mass?

Chemistry
2 answers:
Anna71 [15]3 years ago
7 0

Answer:

Mass, m = 30 grams

Explanation:

It is given that,

Density of an object, d=3\ g/ml

Volume occupies by the silver, V=10\ ml

We need to find the mass of a piece of silver. The mathematical formula for the density is given by :

d=\dfrac{m}{V}

m=d\times V

m=3\ g/ml\times 10\ ml

m = 30 grams

So, the mass of the piece of the silver is 30 grams. Hence, this is the required solution.

ddd [48]3 years ago
4 0
Density = Mass/Volume => Mass = Density*Volume => Mass = 3g/ml*10ml =30g
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Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with
emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

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<em>2Na + 2H₂O → 2NaOH + H₂,</em>

  • The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,
  • Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,

where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,

V is the volume of the gas in L <em>(V = 7.80 L)</em>,

n is the number of moles in mole,

R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,

T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

<em>Using cross multiplication:</em>

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

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