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3 years ago

What is the Molar mass of Na3P

Chemistry

1 answer:

Sedaia [141]3 years ago

6 0

Answer:

99.943071 g/mol

Explanation:

22.98977*3+30.973761

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Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad

ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O

20 mmol 36 mmol 20 mmol

16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO → Na⁺ + HCOO⁻

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So: [OH⁻] = 16 mmol / 400 mL → 0.04 M

- log [OH⁻] = pOH → 1.39

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3 years ago

Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the

Yanka [14]

Answer:

For a: The mixture will need to produce more reactants to reach equilibrium.

For b: The mixture will need to produce more reactants to reach equilibrium.

For c: The mixture will need to produce more products to reach equilibrium.

Explanation:

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

The expression of $K_{p}$ for above equation follows:

$K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$ .....(1)

We are given:

Value of $K_p$ = 0.26

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

For the given options:

For a:

We are given:

$p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For b:

We are given:

$p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For c:

We are given:

$p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm$

Putting values in expression 1, we get:

$Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008$

As, $K_{p}>Q_p$ ; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

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3 years ago

How many sulfur atoms are there in 5.52 mol of sulfur

PSYCHO15rus [73]

Moles * Avogadro constant = 5.52 * 6.022 x 10^23

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