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2 years ago

How many moles are in 1.24 * 10^25 of carbon?

Chemistry

1 answer:

natta225 [31]2 years ago

6 0

Answer:

2.059800664 x 10^47 moles

Explanation:

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Someone PLZ help ASAP

Lera25 [3.4K]

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3 years ago

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Calculate the pH of a 0.2 M * 4 solution for which Kb = 1.8*10^-5 at 26 c . The equation for the reaction

lara31 [8.8K]

<u>Answer:</u> The pH of the solution is 11.24

<u>Explanation:</u>

We are given:

Molarity of ammonia = 0.2 M

$K_b=1.8\times 10^{-5}$

The given chemical equation follows:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

I: 0.2

C: -x +x +x

E: 0.2-x x x

The expression for equilibrium constant follows:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Putting values in above expression, we get:

$1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}$

Neglecting the negative value of x as concentration cannot be negative.

So, $[OH^-]=x=1.88\times 10^{-3}M$

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

$pOH=-\log [OH^-]$

Putting values in above equation, we get:

$pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76$

We know:

$pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24$

Hence, the pH of the solution is 11.24

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3 years ago

Please help I’ll give brainliest

yuradex [85]

Halogens. Ex. fluorine can be the gas,bromine can be the liquid, and iodine could a solid all under room conditions.

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4 years ago

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Water can form a solution by mixing with:

MA_775_DIABLO [31]

Explanation:

Solid, liquids and gases

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3 years ago

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Rank the species CH2NH, CH3I, HF, C2H6 from most to least miscible in ammonia (NH3) 1. CH3I > C2H6 > HF > CH2NH 2. CH2N

777dan777 [17]

Answer:

$HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}$

Explanation:

$NH_{3}$ is an weak base. HF is an weak acid. Hence an acid-base reaction will ocuur between $NH_{3}$ and HF resulting a homogeneous mixture.

$NH_{3}$ is a polar protic molecule. Hence it can form hydrogen bonding with another polar protic molecule $CH_{2}NH$ . Also dipole-dipole attraction force along with london dispersion force exists between $CH_{2}NH$ and $NH_{3}$ .

$CH_{3}I$ is a polar molecule. Hence only dipole-dipole attraction force along with london dispersion force operates between $CH_{3}I$ and $NH_{3}$ .

$C_{2}H_{6}$ is a non-polar molecule. Hence only london dispersion force exists between $C_{2}H_{6}$ and $NH_{3}$ .

So, order of miscibility: $HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}$

(most) (least)

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3 years ago