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Ksenya-84 [330]
2 years ago
14

2.478 g of white phosphorus was used to make phosphine according to the equation: P₄(s) + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(

aq) Calculate the amount, in mol, of white phosphorus
Chemistry
1 answer:
prohojiy [21]2 years ago
3 0

<u>Answer:</u> The amount of white phosphorus is 0.0199 moles

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of white phosphorus = 2.478 g

Molar mass of white phosphorus = 124 g/mol

Putting values in above equation, we get:

\text{Moles of white phosphorus}=\frac{2.478g}{124g/mol}=0.0199mol

Hence, the amount of white phosphorus is 0.0199 moles

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natka813 [3]
930 g for 15 clementines
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If a certain gas occupies a volume of 13 L when the applied pressure is 6.5 atm , find the pressure when the gas occupies a volu
Sav [38]
We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)

Solve for P2 on your calculator and that should get you to the answer.


5 0
2 years ago
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The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters
Ivenika [448]

Answer:-  4.15*10^-^2^0\mu L

Solution:- It is a volume unit conversion problem where we are asked to convert the volume from cm^3 to microliters.

We know that:

1cm^3 = 1 mL

1mL=10^-^3L

and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

= 4.15*10^-^2^0\mu L

So, the answer is  4.15*10^-^2^0\mu L .

7 0
2 years ago
Which of the following shows an accurate combustion reaction?
Jet001 [13]

Answer:

  • <em>Cu + O₂  → CuO₂</em>

Explanation:

A <em>combustion reaction</em> is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

  • <em>Cu + O₂ → CuO₂</em>

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

5 0
2 years ago
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
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