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3 years ago

17.kg into g convert following

Chemistry

1 answer:

vodka [1.7K]3 years ago

6 0

17000 grams and this is a filler

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Which statement about matter is true?

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D. Matter is anything that is made of organic substances and is found on the earth.

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Scientific theories are based on evidence.....<br> True<br> False

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How many moles of iron is needed to react completely with 5.00 moles of sulfur to form iron

Ghella [55]

Answer:

5 moles of Fe(II) are required to react completely with the 5 moles of Sulphur

Explanation:

The balanced equation in this question is

Iron + Sulfur = Iron(II) Sulfide

Fe (II) + S --> Fe(II)S

Thus one mole of sulfur reacts with one mole of Fe(II)

Hence, 5 moles of Fe(II) are required to react completely with the 5 moles of Sulphur

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3 years ago

Dimensional analysis: if you count at the rate of 2 numbers every secound, how many days whould it take you to count to 1 millio

son4ous [18]

Answer: 5 days

Explanation:

1,000,000/2 = 500,000 *do this because of the 2 numbers every second

500,000/60= 8,333.33 *do this to get minutes

8,333.33/60= 138.88 *do this to get hours

138.88/24= 5.7866 *do this to get days

5 days 7 hours 8 minutes 6 seconds

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3 years ago

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌

Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of $CO$ = 0.35 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M$

Initial moles of $H_2O$ = 0.40 mole

Volume of container = 1 L

Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40M 0 0

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}$

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

$K_c=0.70$

Thus the value of the equilibrium constant is 0.70.

5 0

3 years ago