What unit is used to express the average kinetic energy of a system

How many milliliters of a stock solution of 6.20 M HNO3 would you have to use to prepare 0.130 L of 0.470 M HNO3?

Mandarinka [93]

Answer:

9.85

Explanation:

M1V1 =M2V2

6.20×v1= 0.470×0.130

v1 = ( 0.470 × 0.130 ) ÷ 6.20

v1 = 0.0098 L × 1000

V1 = 9.8 ml

6 0

3 years ago

What is the mass number of C12,C13 and C14

Elena L [17]

Carbon-13 and Carbon-14 are isotopes of the element carbon. All carbon atoms have 6 protons in their nucleus and most have 6 neutrons. Since the mass number consists of # of protons + # of neutrons, Carbon 12 has an atomic mass of 12 ( = 6 protons + 6 neutrons). Carbon-13 atoms have one extra neutron, giving it a total of 7 neutrons. Carbon-14 atoms have two extra neutrons, giving them a total of 8 neutrons. Therefore the mass number of C-12= 12, C-13= 13 and C-14= 14.

3 0

4 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Jason is preparing curry from an Indian cookbook his British brother-in-law gee him for his birthday. Unfortunately, the British

Marina CMI [18]

Answer:

0.99 pounds

2,000 oz

0.423 cups

Explanation:

In order to convert this units we need to look up their equivalences.

1 g equals 0,0022 pounds approximately

Then we need to cross-multiply:

$\frac{1 g}{0.0022 pounds} = \frac{450 g}{ Xpounds} \\x = \frac{450 g x 0.0022 pounds}{1 g}$

450 g equals 0.99 pounds approximately

We can do the same calculation for the other 2 ingredients

1 g equals 16 oz

Then (125 g x 16 oz) / 1 oz = 2,000 oz (or 500 4 oz)

1 cup equals 236.59 ml

Then (100 ml x 1 cup)/ 236.59 = 0.423 cups

6 0

3 years ago

Please help. Will give brainliest.

elena-14-01-66 [18.8K]

Answer:

B.

Explanation:

i think its B

7 0

3 years ago