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2 years ago

13

How many grams of solute are needed to order to prepare 100.00 mL of a 0.1000 M solution of a compound with a molecular weight o

f 350.0 g/mol

2 answers:

Troyanec [42]2 years ago

7 0

The grams of solute are required.

The mass of solute is 3.5 g

c = Molarity = 0.1 M

M = Molar mass = 350 g/mol

V = Volume of solution = 100 mL = 0.1 L

n = Number of moles

m = Mass of solute

Molarity is given by

$c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}$

Molar mass is given by

$M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}$

The mass of solute is 3.5 g

Learn more:

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slava [35]2 years ago

4 0

Answer:

i got 35 but i am as lost as you are so don't really take my answer as a trusted answer

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denis-greek [22]

Answer:

The atomic numbers of four elements A, B, C, and D are 6,8,10, and 12 respectively. The two elements which can react to form ionic bonds (or ionic compounds) are B (8= 2,6) and D (12 =2,8,2). So D donates its two electrons to B to fulfill their octet.

Explanation:

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2 years ago

Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el

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Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

$Ni^{2+}+2e^-\rightarrow Ni$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of Nickel

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Given quantity = 18.0 kg = 18000 g (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

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where Q= quantity of electricity in coloumbs = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

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3 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:

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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:

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

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3 years ago

Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)

stepan [7]

Answer:

<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

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<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

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<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

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3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago