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2 years ago

PLESE HELP WILL GIVE BRAINLYESTY

Chemistry

1 answer:

Anastaziya [24]2 years ago

3 0

Answer:

My name is Isaac Newton and I am a scientist. I am known as one of the most important scientists in history. I was born in Woolsthorpe, England on January 4, 1643. My father, a farmer who was also named Isaac Newton, had died three months before my birth. My mother remarried when I was three years old and left me in the care of my grandparents.

Explanation:

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A chemist must dilute 61.9 mL of 548. nM aqueous sodium carbonate (Na2CO3) solution until the concentration falls to 484. nM . H

maw [93]

Answer:

The final volume will be "70.08 mL".

Explanation:

The given values are:

Molar mass,

M1 = 548 nM

or,

= $5.48\times 10^{-7} \ M$

M2 = 484 nM

or,

= $4.84\times 10^{-7} \ M$

Volume,

V1 = 61.9 mL

V1 = ?

By using the expression, we get

⇒ $M1\times V1=M2\times V2$

or,

⇒ $V2=\frac{M1\times V1}{M2}$

By substituting the values, we get

$=\frac{5.48\times 10^{-7}\times 61.9}{4.84\times 10^{-7}}$

$=\frac{339.212}{4.84}$

$=70.08 \ mL$

7 0

2 years ago

What four things affect the tides

S_A_V [24]

I know three
The Moon
The Wind
and Earths Rotation.

Gravity Maybe one thing but i am not sure

4 0

2 years ago

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

2 years ago

Read 2 more answers

Please help. It's for chemistry

irinina [24]

What do i help with what

3 0

2 years ago

how many milliliters of 1.50 m hno3 contain enough nitric acid to dissolve an old copper penny with a mass of 3.94 g?

Lerok [7]

Molar mass HNO₃ = 63.0 g/mol

number of moles = 3.94 / 63.0 => 0.0625 moles

Volume = moles / molarity

V = 0.0625 / 1.50

V = 0.04166 L x 1000 = 41.66 mL

hope this helps!

5 0

2 years ago