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3 years ago

A gas has an initial volume of 15 L. If the temperature increases from 330 K to 450 K, what is the new volume.

Chemistry

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

20.(45)L or about 20.4545L

Explanation:

PV = nRT

Where:

P - pressure

V - volume

n - number of particle moles

R - a constant

T - temperature in K

We can assume the P and n (and definitely R) stay the same, so we infer that

$V_1 = \frac{nRT_1}{P} = 15L\\V_2 = \frac{nRT_2}{P}\\V_1 / V_2 = \frac{nRT_1}{P} / \frac{nRT_2}{P} = \frac{T_1}{T_2}\\\\15L / V_2 = \frac{330K}{450K} = \frac{11}{15}\\\\V_2 = 15L \cdot \frac{15}{11} = 20.(45)L$

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According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra

Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0

3 years ago

When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced

Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields $CO_2$ and $H_2O$

$C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)$

When only 2.5 mol of $O_2$ are consumed in order to complete the reaction, ________ mol of $CO_2$ are produced.

Answer: Thus when 2.5 mol of $O_2$ are consumed in their reaction, 1.5 mol of $CO_2$ are produced

Explanation:

The balanced chemical equation is:

$C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)$

According to stoichiometry :

5 moles of $O_2$ produce = 3 moles of $CO_2$

Thus 2.5 moles of $O_2$ will produce = $\frac{3}{5}\times 2.5=1.5$ moles of $CO_2$

Thus when 2.5 mol of $O_2$ are consumed in their reaction, 1.5 mol of $CO_2$ are produced

4 0

3 years ago

Solid a is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What ty

velikii [3]

This is covalent network type of solid.

For example, silicon dioxide (SiO₂) is covalent network solid with covalent bonding.

Covalent network solid is a chemical compound (or element) in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material.

Silicon(IV) oxide has continuous three-dimensional network of SiO₂ units and diamond has sp3 hybridization.

This solids do not have free electrons so they are good insulators.

They have strong covalent bonds, so they melt at extremely high temperature.

Other examples are quartz, diamond, and silicon carbide.

More about network solid: brainly.com/question/15548648

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3 0

2 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

Explain why the boiling point of H2S with relative molecular mass 34 is lower than that of H2O with relative molecular mass 18

Levart [38]

Answer and Explanation:

In H2O molecules, due to formation of intermolecular hydrogen bonds, there is molecular association. Large amount of energy is required to break these intermolecular hydrogen bonds. Intermolecular hydrogen bonding is not possible in H2S. Hence, its boiling point is lower and is a gas.

4 0

3 years ago