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KengaRu [80]
3 years ago
14

Compute 6.28×1013+7.30×1011.

Chemistry
1 answer:
Simora [160]3 years ago
6 0
6.28×1013+7.30×1011 this =13741.94
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When 250 ml of water is added to 35 ml of 0.2 M HCl
Paladinen [302]
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
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The compound NacI is an example of a/an
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The compound NaCI is an example of a salt. Salt is formed from a neutralization action of an acid and a base. From the type of reaction itself, we can say that the pH should be neutral or at pH 7.0. No matter what type of acid or base is used.
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How to answer this ?
nadezda [96]
You multiply (4×10)by*
6 0
2 years ago
Name three physical properties that differentiate oxides from hydroxide group of minerals.
erik [133]

Answer:

2k20

Explanation:

3 0
2 years ago
The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does t
Mama L [17]

Answer:

6.1%

Explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law  at constant pressure,

\frac{V_1}{T_1} =\frac{V_2}{T_2}

T_2=39\°\ C=39+273=312\ K

T_1=59°\ C=273+59=332\ K

V_2=V_1\times \frac{T_2}{T_1} \\=V_1\times\frac{312}{332}\\=0.939V_1

Percentage change in volume

\%\ change\ in\ volume=\frac{V_1-0.939V_1}{V_1} \times 100\\=6.1\%

Change in volume of the balloon is 6.1%

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3 years ago
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