Mass of CO₂ evolved : 0.108 g
<h3>Further explanation</h3>
Given
1.205g sample, 36% MgCO3 and 44% K2CO3
Required
mass of CO2
Solution
0.36 x 1.205 g=0.4338 g
mass C in MgCO₃(MW MgCO₃=84 g/mol, Ar C = 12/gmol)
= (12/84) x 0.4338
= 0.062 g
0.44 x 1.205 g = 0.5302 g
Mass C in K₂CO₃(MW=138 g/mol) :
= (12/138) x 0.5302
= 0.046 g
Total mass Of CO₂ :
= 0.062 + 0.046
= 0.108 g
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<u>Answer: </u>The correct answer is Silver.
<u>Explanation:</u>
Specific heat of fusion is defined as the amount of heat which is required to raise the temperature of 1 gram of a substance to 1°C. It is generally expressed in kJ/mol
We are required to find the substance which require more heat. For that we need to know the specific heat of all the substances.
The substance which have the highest specific heat, will require more heat.
The specific heat of the given substances are:
Silver = 11.3 kJ/mol
Sulfur = 1.7175 kJ/mol
Water = 5.98 kJ/mol
Lead = 4.799 kJ/mol
The specific heat of silver is the highest and hence, will require more heat.
Hence, the correct answer is silver.
Answer:
50.8 g
Explanation:
Equation of reaction.

From the given information, the number of moles of methane = mass/ molar mass
= 15.4 g / 16.04 g/mol
= 0.960 mol
number of moles of oxygen gas = 90.3 g / 32 g/ mol
= 2.82 mol
Since 1 mol of methane requires 2 moles of oxygen
Then 0.960 mol of methane will require = 0.960 mol × 2 = 1.92 mol of oxygen gas
Thus, methane serves as a limiting reagent.
2.82 mol oxygen gas will result in 2.82 moles of water
So, the theoretical yield of water = moles × molar mass
= 2.82 mol × 18.01528 g/mol
= 50.8 g
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>