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2 years ago

Convert 80 ml to kl.

Chemistry

1 answer:

Mila [183]2 years ago

5 0

Answer:

8e-5.

Explanation:

This is because in order to convert 80 ml to kl, you must divide the volume value by 1e+6.

Hope this helps you, have a great day!

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Determine the electron geometry (eg) and molecular geometry (mg) of <img src="https://tex.z-dn.net/?f=CH_3%5E%2B" id="TexFormula

Yuri [45]

Answer : The correct option is, (e) eg = trigonal planar, mg = trigonal planar

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, $CH_3^+$

$\text{Number of electron pair}=\frac{1}{2}\times [4+3-1]=3$

That means,

Bond pair = 3

Lone pair = 0

The number of electron pair are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.

Hence, the electron geometry (eg) and molecular geometry (mg) of $CH_3^+$ is, trigonal planar and trigonal planar respectively.

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2 years ago

What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?

svp [43]

The answer is 104.9, i dont know how though

4 0

3 years ago

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

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3 years ago

At what the pressure, in atmospheres, that O2, gas at 75°C has a density of 2.00 g/L

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3 years ago

What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?

yawa3891 [41]

The balanced chemical reaction will be:

2H2O = 2H2 + O2

We are given the amount of water used in the decomposition reaction. This will be our starting point.

17.0 g H2O (1 mol H2O/ 18.02 g H2O) (1 mol O2/2 mol H2O) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield =10.2 g / 15.09 g x 100

Percent yield = 67.58%

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2 years ago

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