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2 years ago

What is this please helpme

Chemistry

2 answers:

Helga [31]2 years ago

6 0

It will gain 1 electron

AleksandrR [38]2 years ago

3 0

It will gain one electron

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How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0

3 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

3 years ago

1. A 40,000 kg plane is accelerating at a rate of 4m/s2. How much force does the engine put forth?

mamaluj [8]

Answer:

Explanation:

F=ma

m=40000

a=4

5 0

3 years ago

Which statement defines activation energy?

ehidna [41]

Answer:

The statement which defines activation energy is that it is the difference between reactant energy and maximum energy.J

Explanation:

A reaction that does not have activation energy will not take place at all.

3 0

3 years ago

Audrey is driving 50 miles in 2 hours what is her average speed ?

LenaWriter [7]

It depends on the unit though if its in miles per hour then you can simply divide 50 by 2

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3 years ago