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inessss [21]
3 years ago
10

1) Given 1469 of H₂SO4 how many moles do you have

Chemistry
1 answer:
mart [117]3 years ago
4 0

Answer:

15moles

Explanation:

Given parameters:

Mass of H₂SO₄   = 1469g

Unknown:

Number of moles  = ?

Solution:

To solve this problem, let us first find the molar mass of the given compound;

  molar mass = 2(1) + 32 + 4(16)  = 98g/mol

So;

 Number of moles = \frac{mass}{molar mass}    = \frac{1469}{98}  

 Number of moles  = 15moles

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ikadub [295]

Answer:

I think the answer is gravity

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3 years ago
Determine the percent composition by mass of a 100g salt solution which contains 20g salt
prisoha [69]

Answer:

Mass of solution=100g

mass of salt=20g

so; mass of solute=80g

percentage composition =(mass of salt/total

mass) ×100

=  \frac{20}{100}  \times 100 \\  = 20\%

glad to help you

hope it helps

8 0
3 years ago
Read 2 more answers
Find the ph of of 100 ml of an aqueous 0.43m baoh2 solution
denpristay [2]
Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.
4 0
3 years ago
If gas is truly to be classified as a form of matter this means it can be measured. What form of gas can be measured?
juin [17]

Answer:

Gas is sometimes measured in cubic feet at a temperature of 60 degrees Fahrenheit and an atmospheric pressure of 14.7 pounds per square inch. Gas production from wells is discussed in terms of thousands or millions of cubic feet (Mcf and MMcf). Resources and reserves are calculated in trillions of cubic feet (Tcf).

3 0
3 years ago
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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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