<span>calcium is not soluble in water because it reacts to water. It does not dissolve in it.
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The equation is x/25=3.2
*25 *25
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x=80g
Answer is in picture below.
Use 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; mass percentage of the chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; amount of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; amount of carbon.
n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.
This compound is dichlorocarbene CCl₂.
Question A is more testable through a scientific experiment because you have a specific thing you are trying to find out, "which gelatin takes longer to set".
Answer:
C₂H₄O
Explanation:
In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.
<em>Moles CO₂ = Moles C:</em>
11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =
3.216g C
<em>Moles H₂O = 1/2 moles H:</em>
4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =
0.537 mol H * (1g/mol) = 0.537g H
<em>Mass O to find moles O:</em>
5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O
<em>Ratio of atoms -Dividing in 0.134 moles-:</em>
C = 0.268mol C / 0.134 mol O = 2
H = 0.537mol H / 0.134 mol O = 4
O = 0.134mol O / 0.134 mol O = 1
Empirical formula is:
<h3>C₂H₄O</h3>
