All categories

3 years ago

The temperature of a sample of gas in steel container at 30.3 Atm is

Chemistry

2 answers:

vladimir1956 [14]3 years ago

6 0

Answer:

74.09 atm

Explanation:

Using the gas laws ( Charles and Boyle's law). We have the formula ,

P1/T1 = P2/T2

Where P1 = 30.3atm

T1 = -100 degree Celsius

to kelvin = -100+ 273 = 173K

T2 = 150 degree Celsius

To Kelvin = 150 = 150+273 = 423K

Imputing values

P1/T1 = P2/T2

30.3/173 = P2/ 423

Cross multiply

173×P2 = 30.3 ×423

173P2 = 12816.9

Divide both sides by 173

P2 = 12816.9/173

P2 = 74.09 atm

I hope this was helpful, please mark as brainliest

ddd [48]3 years ago

6 0

Answer:

P2=74.086atm

Explanation:

Using the formula p1/t1=P2/t2

P1=30.3atm

P2=?

T1=-100°c to Kelvin=273-100=173k

T2=150°c to Kelvin=273+150=423k

30.3\173=x/423

Cross multiply

173x=423×30.3

173x=12816.9

Divide both sides by 173

X=74.086atm

You might be interested in

Where can you find the number of electrons of an<br> element?

andrezito [222]

Answer:

Subtract the charge from the atomic number. When an ion has a positive charge, the atom has lost electrons. To calculate the remaining number of electrons, you subtract the amount of extra charge from the atomic number. In the case of a positive ion, there are more protons than electrons.

Explanation:

5 0

3 years ago

Read 2 more answers

A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to

Drupady [299]

Answer : The final temperature of the mixture is $29.6^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = $0.499J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron = 39.9 g

$m_2$ = mass of water = $Density\times Volume=1g/mL\times 50.0mL=50.0g$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of iron = $78.1^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC$

$T_f=29.6^oC$

Therefore, the final temperature of the mixture is $29.6^oC$

8 0

3 years ago

Which would be attracted most to a magnet?

Debora [2.8K]

Either iron or metal

8 0

3 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?

GenaCL600 [577]

K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent

KNO₃ is limiting reagent

5 0

3 years ago