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3 years ago

Help i’ll mark brainliest asap please

Chemistry

2 answers:

agasfer [191]3 years ago

5 0

Answer:

The 2nd One.

Explanation:

The skin and the skin cells were active in Norman's experiment until he seperated them in which the cyptoplasm then became active.

12345 [234]3 years ago

4 0

2nd one is the right answer

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Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant o

Zina [86]

Answer:

The equilibrium concentrations of all species :

$[H_2]= 0.314 M$

$[F_2]=0.314 M$

$[HF]=3.372 M$

Explanation:

Moles of hydrogen gas = 3.00 mole

Concentration of hydrogen gas = $[H_2]=\frac{3.00 mol}{1.5 L}=2.0 M$

Moles of fluorine gas = 3.00 mole

Concentration of fluorine gas = $[F_2]=\frac{3.00 mol}{1.5 L}=2.0 M$

Given the equilibrium constant of the reaction = $K_c=1.15\times 10^2$

$H_2(g)+F_2(g)\rightleftharpoons 2HF(g)$

Initial

2.0 M 2.0 M 0

At equilibrium

(2.0-x)M (2.0 -x)M 2x

An expression of $K_c$ is given by :

$K_c=\frac{[HF]^2}{[H_2][F_2]}$

$1.15\times 10^2=\frac{(2x)^2}{(2.0-x)(2.0 -x)}$

On solving for x , we get :

x = 1.686 M

The equilibrium concentrations of all species :

$[H_2]=(2.0 -1.686)M = 0.314 M$

$[F_2]=(2.0 -1.686)M = 0.314 M$

$[HF]=(2\times 1.686)M = 3.372 M$

8 0

3 years ago

Can some one tell me the best position in sex

masya89 [10]

Answer:

speedbump and magician

Explanation:

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3 years ago

What is the hydrogen ion concentration of a solution with ph = 2. 25?.

svetoff [14.1K]

6x10-3M that will be your answer muh friend yes i know i know im big brain

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2 years ago

Which side does the light always start from?

jarptica [38.1K]

Answer:

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2 years ago

If a solution containing 51.429 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 16.642 g

OLga [1]

Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass

moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles

moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4

Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles

mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams

7 0

4 years ago