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Leya [2.2K]
3 years ago
7

When mowing the lawn what greenhouse gas are you generating

Chemistry
1 answer:
trasher [3.6K]3 years ago
4 0
You are createing carbon dioxide
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Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is
Citrus2011 [14]
Depression in freezing point (ΔT_{f}) = K_{f}×m×i,
where, K_{f} = cryoscopic constant = 1.86^{0} C/m,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For NaNO_{3})

Thus, (ΔT_{f}) = 1.86 X 0.0085 X 2 = 0.03162^{0}C

Now, (ΔT_{f}) = T^{0} - T
Here, T = freezing point of solution
T^{0} = freezing point of solvent = 0^{0}C
Thus, T = T^{0} - (ΔT_{f}) = -0.03162^{0}C
8 0
3 years ago
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Which choice below would match homozygous recessive?
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The answer is B

Homozygous mean the same and you know that the two alleles will be the same (either BB or bb) and receive is usually the lower case set of alleles
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Answer:

1. They lack a nucleus

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3 years ago
Wood, Iron, Glass, Paper, Nickel Which substances show magnetic property?
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If I am correct the answer would be iron and nickel.
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3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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