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1 year ago

What is the percent oxygen in something made of 2 mole of copper and 1 mole of oxygen called copper (l) oxide?

Chemistry

1 answer:

Lady_Fox [76]1 year ago

8 0

Answer:

20.114%

Explanation:

Percent composition by element Element Symbol Mass Percent Copper Cu 79.886% Oxygen O 20.114%

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Which element has the strongest attraction for electrons?

NemiM [27]

Answer:

The answer is Flourine

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2 years ago

A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u

Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

$Molarity=\frac{n\times 1000}{V_s}$

where,

Morality = 0.612 M

n= moles of solute

$V_s$ = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

$0.612=\frac{n\times 1000}{100ml}$

$n=0.0612moles$

$Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g$

Therefore, the mass of copper (II)nitrate required is 11.5 grams

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2 years ago

Freé point<br><img src="https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20" id="Te

gladu [14]

Answer

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2 years ago

Read 2 more answers

What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?

zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

V2 = 500 mL

V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL

3 0

2 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

2 years ago