Answer: A. the rotational period of the earth is the same as that of the moon
Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
n - amount of substance.
Answer:
We need 375 milliliters of 0.100 M NaHCO3 solution
Explanation:
Step 1: Data given
Initial molarity NaHCO3 = 0.100 M
Volume prepared solution = 750.0 mL
Molarity prepared solution = 0.05 M
Step 2: Calculate initial volume
C1*V1 = C2V2
⇒with C1 = the initial concentration = 0.100 M
⇒with V1 = The initial volume = TO BE DETERMINED
⇒with C2 = the new concentration = 0.0500M
⇒with v2 = the new volume = 750.0 mL = 0.750 L
0.100 M * V1 = 0.0500 M * 0.750 L
V1 = (0.0500M * 0.750L)/0.100 M
V1 = 0.375 L = 375 mL
We need 375 milliliters of 0.100 M NaHCO3 solution
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>
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