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Which statement best explains why Earth's outer core is in liquid form?

erik [133]

Answer:

My answer im guessing is "d"

Explanation:

5 0

2 years ago

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What is the SI unit for the mass of subatomic particles?

gogolik [260]

hey, im jordan :)

the SI unit for the mass of subatomic particles is amu (atomic mass unit)

hope this helps!

have a great day :D

4 0

2 years ago

In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the

notka56 [123]

Answer:

$P_T=112.4torr$

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

$P_{H_2}=x_{H_2}*P_T$

And can be solved for the total pressure as follows:

$P_T=\frac{P_{H_2}}{x_{H_2}}$

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

$x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667$

Then, we can plug in to obtain the total pressure:

$P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr$

Regards!

4 0

2 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0

2 years ago

Is chemistry hard in grade 11 explain to me everything plsss thanks

damaskus [11]

Answer: yes it’s hard

Explanation:

8 0

2 years ago

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