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BartSMP [9]
3 years ago
7

Which of the following is not the same as 12.5 millimeters? 0.0000125 km 0.00125 hm 0.0125 m 1.25 cm

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
5 0
1 mm (millimeter) = 0.000001 km (kilometer)
12.5 mm = <span>0.0000125 km

1 mm = </span><span>0.00001 hm (hectometer)
12.5 mm = </span><span>0.000125 hm

1 mm = </span>0.001 m (meter)
12.5 mm = 0.0125 m

1 mm = 0.1 cm (centimeter)
12.5 mm = 1.25 cm

So the only one of the answer choices that doesn't equal 12.5 mm is 0.00125 hm, since 12.5 mm is <span>0.000125 hm.

Answer:
</span><span>0.00125 hm
</span><span>
Hope this helps!</span>
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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
2. What group is Ballardium located? (Bu) *
elena-14-01-66 [18.8K]

Answer:

This question appears incomplete

Explanation:

There is no such element known as "Ballardium (Bu)" in the periodic table. However, there are elements with a bit of similarity in spellings and pronunciation such as Beryllium (Be) which is found in group 2 (meaning it is an alkali earth metal), Berkelium (Bk) which is an actinide (meaning it is radioactive) and Vanadium (V) which is found in group 5 of the periodic table (meaning it's a transition metal).

5 0
3 years ago
Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m
devlian [24]

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n  of S = 2/32 = 0.0625kmols

Reaction:  S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

5 0
3 years ago
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masya89 [10]
Answer: Option (D) 30N

Detailed Solution:
According to Newton's second law:

F = ma --- (A)

Given:
mass = 5kg
acceleration = 6 m/s^2
F = ?

Plug all the value in equation (A)
F = (5)(6)
Ans: F = 30N
4 0
3 years ago
Read 2 more answers
Will mark as Brainliest.
SVETLANKA909090 [29]
<span>If I done the math correctly it is 3729J because you multiply 16.5 g  by the 2260 J/g and get 3729 J</span>
4 0
3 years ago
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