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3 years ago

The person who follows me first will be marked branliset and will be followed by me

Chemistry

2 answers:

grandymaker [24]3 years ago

8 0

I follow you
Hi what’s up

nikitadnepr [17]3 years ago

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Answer:

sorry

Explanation:

ya no thanks

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Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a

Dvinal [7]

Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 100.0 atm

$P_2$ = final pressure of gas at STP = 1 atm

$V_1$ = initial volume of gas = 50.0 L

$V_2$ = final volume of gas at STP = ?

$T_1$ = initial temperature of gas = $27.0^oC=273+27.0=300K$

$T_2$ = final temperature of gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}$

$V_2=4550L$

Therefore, the volume of hydrogen gas at STP is 4550 L.

3 0

3 years ago

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri

Nimfa-mama [501]

Answer: The expression for equilibrium constant is $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as $k_c$

For a general reaction:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is written as:

$k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Chemical reaction for the formation of ammonia is:

$N_2+3H_3\rightleftharpoons 2NH_3$

$k_c=1.6\times 10^2$

Expression for $k_c$ is:

$k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

8 0

3 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

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3 years ago

A red blood cell placed in saline solution will shrink. True or false? Explain please?

Scorpion4ik [409]

Yes a red blood cell placed in a sline solution shrinks because of the process of osmosis.

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3 years ago

Which statement correctly describes metallic bonds?

Black_prince [1.1K]

Answer:

Explanation:

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3 years ago

The person who follows me first will be marked branliset and will be followed by me​

The person who follows me first will be marked branliset and will be followed by me