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3 years ago

Question 1

Chemistry

1 answer:

klasskru [66]3 years ago

4 0

Answer:

it will be C

Explanation:

PLS MARK ME AS BRAINLY

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You have a 3.0 g of solid magnesium metal in 250 ml of 1.0mol/l hydrochloric acid solution that is in the beaker. you place a la

k0ka [10]

Answer:

a)Mg(s) + 2HCl(aq) -------> MgCl2(aq) + H2(g)

b) 0.125 moles of hydrogen gas is reduced in the reaction.

C) 3.18 L

d)2.25 g of water

Explanation:

a) the equation of this reaction is;

Mg(s) + 2HCl(aq) -------> MgCl2(aq) + H2(g)

Number of moles= concentration × volume= 1.0 × 250/1000 = 0.25moles of HCl

From the equation;

2 moles of HCl yields 1 mole of hydrogen gas

Hence 0.25 moles of HCl yields 0.25 × 1/2 = 0.125 moles of hydrogen gas

Thus 0.125 moles of hydrogen gas is reduced in the reaction.

P= 760 mmHg (standard pressure)

V= ????

T= 298 K

n= 0.125 moles

R= 0.082 atm dm-3K-1mol-1

Since the gas is collected over water, SVP of hydrogen at 25°c is 28mmHg

Therefore; P=760-28= 732mmHg

But

1 atm =760 mmHg

Therefore 732 mmHg= 732/760= 0.96 atm

PV=nRT

V= nRT/P

V= 0.125 × 0.082 × 298/0.96

V= 3.18 L

Note 1dm-3=1L

2H2(g) + O2(g) ----> 2H2O(g)

From the equation;

2 moles of hydrogen yields 2 moles of water

0.125 moles of hydrogen yields 0.125 moles of water

Mass of water = 0.125 moles × 18gmol-1 = 2.25 g of water

6 0

3 years ago

Will mark brainliest!

aniked [119]

The decoders of the DNA messege

4 0

4 years ago

Draw the mechanism for the williamson ether reaction using m-cresol and benzyl bromide (use sodium methoxide as the base).

Nina [5.8K]

Below is the reaction of ether formation.

3 0

4 years ago

How does examining past data help prepare people for feature weather hazards?

klemol [59]

Answer:

It lets people prepare for future weather hazards:

- If, in certain months and on certain dates, the past data shows that there's a history of rain or heat stroke on those days, people can prepare in the future for those events.

- They can also expect wind speeds, temperatures and stuff like that!

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3 years ago

What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11

PilotLPTM [1.2K]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$CaF_2\rightleftharpoons Ca^{2+}+2F^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ca^{2+}][F^-]^2$

We are given:

$K_{sp}=3.4\times 10^{-11}$

Putting values in above equation, we get:

$3.4\times 10^{-11}=(s)\times (2s)^2\\\\3.4\times 10^{-11}=4s^3\\\\s=2.04\times 10^{-4}mol/L$

To calculate the solubility in g/L, we will multiply the calculated solubility with the molar mass of calcium fluoride:

Molar mass of calcium fluoride = 78 g/mol

Multiplying the solubility product, we get:

$s=2.04\times 10^{-4}mol/L\times 78g/mol=159.12\times 10^{-4}g/L=0.015g/L$

Hence, the correct answer is Option b.

3 0

3 years ago