Ask question

Ask question

All categories

2 years ago

9

How many moles of each element are in one mole of Be(OH)2?

2 answers:

REY [17]2 years ago

8 0

B. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen

Papessa [141]2 years ago

4 0

How many moles of each element are in one mole of Be(OH)2?

1 mole of beryllium, 1 mole of oxygen, 2 moles of hydrogen

1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen CORRECT ANSWER

2 moles of beryllium, 2 moles of oxygen, 2 moles of hydrogen

2 moles of beryllium, 1 mole of oxygen, 1 moles of hydrogen

I took the test and B) was correct

You might be interested in

What is matter?

belka [17]

A Anything that has mass and takes up space

4 0

3 years ago

Read 2 more answers

In your own words, explain the periodic law.

mote1985 [20]

States that the properties of elements are periodic or recurring and are correlated to their atomic number.

6 0

3 years ago

Which mineral might scratch the mineral fluorite, but would not scratch the mineral amphibole? 1 brucite 2. magnesite 3. carnall

NARA [144]

Answer:

olivine i think

Explanation:

4 0

2 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

2 years ago

Conservation of mass was discussed in the background. Describe how conservation of mass (actual, not theoretical) could be check

Sedbober [7]

Answer:

Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:

1. Weigh all the equipment and materials required in the experiment before the experiment.

2. Avoid spillage and evaporation during the experiment.

3. Weigh all the equipment and materials after the experiment.

If the mass is conserved then weight from step 1 is equal to weight from step

3 0

3 years ago