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dsp73
3 years ago
10

The boiling point of a substance in City A is found to be 145°C the boiling point of th same substance in Xity B is 141°C which

city A,or B is at higher elevation and how?
Chemistry
1 answer:
DochEvi [55]3 years ago
4 0
City B. Higher altitudes have lower boiling points due to lower atmospheric pressure at higher altitudes
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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

K cancelled by K.

By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

4 0
3 years ago
PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP
Elza [17]

Answer: There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles

1 mole of H_2O contains =  6.023\times 10^{23} molecules

Thus 3.5 moles of H_2O contains =  \frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23} molecules.

There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

3 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
Which equation represents a neutralization reaction?
abruzzese [7]
C. HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O
8 0
3 years ago
9. Carbon-14 has a half-life of 5,730 years. How many years have passed after 3<br> half-lives?
ella [17]

Answer:

120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300

Explanation:

Hope this helps!

7 0
3 years ago
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