Answer:
Burning wood
Explanation:
the fire releases heat into the air from the burning wood
The answer is a because its not low ductile or gasst
To solve this problem, we must assume ideal gas behaviour so
that we can use Graham’s law:
vA / vB = sqrt (MW_B / MW_A)
where,
<span>vA = speed of diffusion of A (HBR)</span>
vB = speed of diffusion of B (unknown)
MW_B = molecular weight of B (unkown)
MW_A = molar weight of HBr = 80.91 amu
We know from the given that:
vA / vB = 1 / 1.49
So,
1/1.49 = sqrt (MW_B / 80.91)
MW_B = 36.44 g/mol
Since this unknown is also hydrogen halide, therefore this
must be in the form of HX.
HX = 36.44 g/mol , therefore:
x = 35.44 g/mol
From the Periodic Table, Chlorine (Cl) has a molar mass of
35.44 g/mol. Therefore the hydrogen halide is:
HCl
You need the set of reactions that goes from ammonia to nitric acid.
<span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
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Answer:
The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol
Explanation:
Fractional distillation is a separation process based on difference in boiling point of two compounds.
1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.
1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.
Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.
So, boiling point of 1-butanol is much higher than 1-chlorobutane.
Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.
So, option (D) is correct.
