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stellarik [79]
3 years ago
10

What volume of gold would be equal in mass to a piece of copper with a volume of 101 ml? the density of gold is 19.3 g/ml; the d

ensity of copper is 8.96 g/ml?
Chemistry
2 answers:
Kaylis [27]3 years ago
6 0

Answer:

We need 46,88 ml of gold to have the same mass as 101 ml of copper

Explanation:

First we have to calculate the mass of the piece of copper, remember that ml is a measure for volume, so we just have to solve the formula of density for mass:

Density=mass/volume

Mass=volume*density

And we instert the data that we knwo:

Mass=101*8.96

Mass= 904,96 grams

SO know that we now the mass of the copper piece, we can solve for volume in the formula of density to get how many mililiters of gold we need to have 904,96 grams:

Density=mass/volume

Volume=Mass/density

Volume= 904,96/19,3

Volume=46,88 mililiters

Ahat [919]3 years ago
3 0
Answer is: volume of gold  equal in mass to a piece of copper is 46,89 ml.
d(Cu) = 8,96 g/ml.
V(Cu) = 101 ml.
d(Au) = 19,3 g/mol.
V(Au) = ?.
m(Cu) = d(Cu) · V(Cu)
m(Cu) = 8,96 g/ml · 101 ml
m(Cu) = 905 g.
m(Au) = m(Cu) = 905 g.
V(Au) = m(Au) ÷ d(Au)
V(Au) = 905 g ÷ 19,3 g/ml.
V(Au) = 46,89 ml.

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The solubility of copper(i) chloride is 3.91 mg per 100.0 ml of solution. calculate ksp for cucl (cucl=99.00 g mol-1).
garri49 [273]
Convert  Mg  to  grams
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3 0
3 years ago
Read 2 more answers
When 8.0 g H₂ react with 8.0 g O₂ in the reaction 2H₂ + O₂ → 2H₂O, what are the theoretical yield and the limiting reactant?
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Answer:

Now, we have to determine the limiting reagent.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g of

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.

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Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.But 29 g is the given amount. Amount of oxygen unreacted = 29 - 24 = 5 g

7 0
3 years ago
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