The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
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Explanation:
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
Answer:
A. 0.0440 moles/day
Explanation:
First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:
33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag
Now, let's divide this by 7 to get the rate per day:
0.3081 mol Ag ÷ 7 days = 0.0440 mol/day
Thus, the answer is A.
Answer:
Explanation:
Hello there!
In this case, according to the Dalton's law, which explains that the total pressure of a gaseous system equals the sum of the partial pressures of the gases composing, for the gaseous mixture composed by oxygen, nitrogen and carbon dioxide it would be possible to write:
Now, given the pressure of the system and those of oxygen and nitrogen, we calculate that of carbon dioxide as shown below:
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