Answer:
V₂ = 16.5 L
Explanation:
To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:
V₁/n₁ = V₂/n₂
In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].
- n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol
<em>We use the reaction to calculate n₂</em>:
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
0.1900 mol CO * 
0.1900 mol CO₂
0.1900 mol NO * 
0.095 mol N₂
- n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol
Calculating V₂:
22.0 L / 0.3800 mol = V₂ / 0.2850 mol
V₂ = 16.5 L
The formula that we will be using is:
total salt / total solution = .10
Let a be the ml of the first solution
And let b be the ml of the second solution
.25a + .05b / a + b = .10
Also:
a + b = 1400
a = 1400 - b
So substitute:
.25 x (1400 - b) + .05b / 1400 = .10
350 - .25b + 0.05b/ 1400 = .10
350 - .2b / 1400 = .10
350 - .2b = .10 x 1400
.2b = 350 - 140
.2b = 210
b = 1050
a = 1400 - b
a = 1400 - 1050
= 350 mL
Step 1
<em>The reaction involved:</em>
CO + 2 H2 → CH3OH (completed and balanced)
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Step 2
<em>Data provided:</em>
19.7 g H2 (the limiting reactant)
Excess reactant = CO
144.5 g CH3OH = actual yield
----
<em>Data needed:</em>
The molar masses of:
H2) 2.00 g/mol
CH3OH) 32.0 g/mol
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Step 3
The theoretical yield:
By stoichiometry,
CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)
2 x 2.00 g H2 --------- 32.0 g CH3OH
19.7 g H2 --------- X
X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2
X = 157.6 g CH3OH (The theoretical yield)
-----------
Step 4
The % yield is defined as follows:
Answer: d. 93% (it is the nearest value in comparison to my result)
<span>C. C6H12O6
</span>This compound is covalent, and all salts are ionic compounds.
Answer:
Explanation:
The mechanical properties of a material affect how it behaves as it is loaded. The elastic modulus of the material affects how much it deflects under a load, and the strength of the material determines the stresses that it can withstand before it fails
