1) Write the balanced equation to state the molar ratios:
<span>3H2(g) + N2(g) → 2NH3(g)
=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3
What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
First, convert the 250.0 L of NH3 to number of moles at STP .
Use the fact that 1 mole of gas at STP occupies 22.4 L
=> 250.0 L * 1mol/22.4 L = 11.16 L
Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3
=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2
Third, convert 5.58 mol N2 into liters at STP
=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters
Answer: 124,99 liters
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:
2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2
Second, convert the number of moles to liters of gas at STP:
3.75 mol * 22.4 L/mol = 84 liters of H2
Answer: 84 liters
</span>
Answer:
1. CaCO3 + 2HCl → CaCl2 + H2O + CO2
2. C6H12O2 + 8O2 → 6CO2 + 6H2O
Explanation:
Answer: The molecular mass of this compound is 131 g/mol
Explanation:
Depression in freezing point:
where,
= depression in freezing point = 
= freezing point constant = 
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute = 0.49 g
= mass of solvent (cyclohexane) = 20.00 g
= molar mass of solute = ?
Now put all the given values in the above formula, we get:
Therefore, the molar mass of solute is 131 g/mol
Answer:
Explanation:Artificial selection is distinct from natural selection in that it describes selection applied by humans in order to produce genetic change. When artificial selection is imposed, the trait or traits being selected are known, whereas with natural selection they have to be inferred. In most circumstances and unless otherwise qualified, directional selection is applied, i.e., only high-scoring individuals are favored for a quantitative trait. Artificial selection is the basic method of genetic improvement programs for crop plants or livestock (see Selective Breeding). It is also used as a tool in the laboratory to investigate the genetic properties of a trait in a species or population, for example, the magnitude of genetic variance or heritability, the possible duration of and limits to selection, and the correlations among traits, including with fitness.
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
