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astraxan [27]
4 years ago
12

A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperatu

re must this sample be heated?
Chemistry
1 answer:
Anastaziya [24]4 years ago
7 0

Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

Temperature before heating = 35.0°C = 308 K

Volume after heating = 5.25 L

Pressure is constant

Step 2: Calculate temperature

V1 / T1 = V2 /T2

⇒ V1 = the initial volume = 2.00 L

⇒ T1 = the initial temperature = 308 K

⇒ V2 = the final volume = 5.25 L

⇒ T2 = The final temperature = TO BE DETERMINED

2.00L / 308.0 = 5.25L / T2

T2 = 5.25/(2.00/308.0)

T2 = 808.5 K

The sample will be heated to 808.5 Kelvin

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25 L of a gas is collected at 115 kPa. If the pressure increases to 300 kPa, what is the new volume?
Lina20 [59]

Answer:

V2 = 9.58 Litres.

Explanation:

Given the following data;

Initial volume = 25 L

Initial pressure = 115 kPa

Final pressure = 300 kPa

To find the new volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

115 * 25 = 300*V_{2}

2875 = 300*V_{2}

V_{2} = \frac {2875}{300}

V_{2} = 9.58

V2 = 9.58 L

Therefore, the new volume is 9.58 litres.

5 0
3 years ago
Consider this equation: 2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?
weqwewe [10]

Answer:

7

Explanation:

2.524g(5.1)(106)g

(6.85)(103)g

=

1364.4744g2

705.55g

=

1364.4744g

705.55

=

1364.4744g

705.55

=1.933916g

6 0
3 years ago
Another chem question I need help with (multiple choice)
Andreyy89

Answer:

The answer to your question is letter B, 2-methylhexane.

Explanation:

Remember that for naming organic compounds first, we need to look for the largest chain of carbons.

In your example, the largest chain is horizontal and has 6 carbons.

Later, we need to circle all the branches, in your example there is only one branch located close to the left side

After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.

Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.

7 0
3 years ago
Many insulating materials contain spaces filled with air because air is what type of material?
Rina8888 [55]
<span>It is because air is a poor conductor of heat. Good luck ;)</span>
6 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
3 years ago
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