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Leno4ka [110]
3 years ago
12

A 4.5 M solution is to be diluted to 750.0 mL of a 1.5 M solution. How many mL of the 4.500 M solution are required?

Chemistry
1 answer:
maw [93]3 years ago
7 0

Answer:

309

Explanation:

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If a gas has a volume of 350 mL at 780 mm Hg, what pressure will be needed to cause its volume
makkiz [27]

The required pressure of the gas is 546 mmHg.

<h3>What is the relation between volume and pressure?</h3>

Relation between the volume and pressure of gas is explained through the ideal gas equation PV = nRT, and for the question required equation is:

P₁V₁ = P₂V₂, where

P₁ & V₁ are the initial pressure and volume.

P₂ & V₂ are the final pressure and volume.

On putting values from the question to the equation, we get

P₂ = (780)(350) / (500) =

P₂ = 546 mmHg

Hence required pressure is 546 mmHg.

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5 0
2 years ago
Urea is a common fertilizer with the formula CO(NH2)2 that is sometimes used as a chemical de-icer for icy road surfaces in the
nordsb [41]

Answer:

The answer to your question is  molality = 0.61

Explanation:

Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.

Data

Kf = 1.86 °C/m

molality = ?

ΔTc = 1.13°C

Formula

ΔTc = kcm

Solve for m

m = ΔTc/kc

Substitution

m = 1.13 / 1.86

Simplification and result

m = 0.61

4 0
4 years ago
Analyze the given diagram of the carbon cycle below Part 1: Which compound does C represent? Part 2: Name a process that could r
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1: CO2 cycle

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3: in soil repiration grass regenerates and is ready for gazing, the carbon will be sequestered again when the animals return. then CO2 is released into the atmosphere.

5 0
3 years ago
To be able to see a measurable effect, your boiling point elevation must be at least 1.00°C. Knowing that the Kb­ of water is 0.
ale4655 [162]

Answer : The mass of calcium chloride (in g) needed is, 1.92 grams.

Explanation : Given,

Boiling point of elevation constant (K_b) for water = 0.512^oC/m

Mass of water (solvent) = Density\times Volume=1.00g/mL\times 26.63mL=26.63g=0.02663kg

Molar mass of CaCl_2 = 110.98 g/mole

Formula used :  

\Delta T_b=i\times K_b\times m\\\\\Delta T_b=i\times K_b\times\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2\times \text{Mass of water in kg}}

where,

\Delta T_b = change in boiling point  = 1.00^oC

i = Van't Hoff factor = 3 (for electrolyte)

K_b = boiling point constant for water

m = molality

Now put all the given values in this formula, we get

1.00^oC=3\times (0.512^oC/m)\times \frac{\text{Mass of }CaCl_2}{110.98g/mol\times 0.02663kg}

\text{Mass of }CaCl_2=1.92g

Therefore, the mass of calcium chloride (in g) needed is, 1.92 grams.

4 0
3 years ago
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