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3 years ago

_NH3 + _ 02 →_NO + _ H2O Balanced

Chemistry

1 answer:

vovikov84 [41]3 years ago

8 0

Answer:

4NH3+5O2=4N0+6H2O

Explanation:

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3 years ago

Is that the right answer

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Is what the right answer

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3 years ago

concentrated phosphoric acid is90% H3PO4 by mass and the remaining mass is water. The molarity of H3PO4 is 12.2M at temperature

blondinia [14]

Answer:

82.0 mL

Explanation:

Step 1: Given data

Concentration of concentrated acid (C₁): 12.2 M

Volume of concentrated acid (V₁): ?

Concentration of dilute acid (C₂): 1.00 M

Volume of dilute acid (V₂): 1.00 L

Step 2: Calculate the required volume of the concentrated acid

We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL

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3 years ago

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3 years ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

i: van 't Hoff factor (1 for non-electrolytes)
Kf: cryoscopic constant
m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

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3 years ago