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aleksandr82 [10.1K]
3 years ago
10

If 1000. mL of water freezes, which of the following is a reasonable approximation for the volume of the resulting ice?

Chemistry
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

If 1000. mL of water freezes, which of the following is a reasonable approximation for the volume of the resulting ice?

Group of answer choices

1000. mL

961 mL

1040 mL

Explanation:

Ice is fewer denser than water.

The reason is the volume occupied by the same mass of ice with water is more than the volume occupied by water. Ice has more empty space within it.

Due to this reason, ice floats on water.

When 1000ml of water freezes to ice then its volume is greater than water.

Among the given options the correct answer is 1040 mL .

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If 0.50 g of O2(g) reacts with excess H2(g), what is the volume of H2O(g) obtained from the reaction at STP?
ozzi

Answer:

0.7 L H2O

Explanation:

4 0
2 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
The muscles of the body are part of the muscular system but would not operate without the _______ system providing the impulses
MatroZZZ [7]

Answer:

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5 0
3 years ago
Iron reacts with oxygen to produce iron oxide rust this reaction is representative in an equation as to FE plus XO22FE203 identi
igomit [66]

The balanced equation :

4Fe + 3O₂⇒ 2Fe₂O₃

<h3>Further explanation</h3>

Given

Rust reaction

Required

Balanced equation

Solution

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Fe + O₂⇒ 2Fe₂O₃

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In the above reaction :

Fe, left=1, right = 4, so coefficient in the left=4

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3 0
2 years ago
You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca
defon
The equation to be used are:

PM = ρRT
PV = nRT
where
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The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
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Solving for V,
V = 4.839×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.839×10⁻⁴ m³ * 1000
V = 0.484 L
5 0
3 years ago
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