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2 years ago

Write a sentence for the following equation: HCI (aq) + NaOH(aq) --> NaCl (aq) + H2O(l)

Chemistry

1 answer:

DedPeter [7]2 years ago

5 0

Explanation:

it shows the Neutralization process, Acid (HCl) reacts with base(NaOH) to give salt(NaCl) and water (H2O)

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Which of the following bonds is the most reactive? A. C—C B. N—H C. O—O D. C—H

USPshnik [31]

I think the answer is b

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2 years ago

For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- what species are a conjugate acid-base pair?

aleksklad [387]

Ionic equation:

$\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}$

The acid and base in a conjugate pair differ by only one proton $\text{H}^{+}$ . The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

$\text{H}_2\text{PO}_4^{-}$ loses one proton to produce $\text{HPO}_4^{2-}$ in this reaction.

$\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}$

Meanwhile, $\text{HAsO}_4^{2-}$ gains one proton to form $\text{H}_2\text{AsO}_4^{-}$ .

$\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}$

Therefore

$\text{H}_2\text{PO}_4^{-}$ is the conjugate acid $\text{HPO}_4^{2-}$ , its conjugate base.
$\text{HAsO}_4^{2-}$ is the conjugate base of $\text{H}_2\text{AsO}_4^{-}$ , its conjugate acid.

8 0

3 years ago

Read 2 more answers

When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef

kumpel [21]

Answer:

molar mass of the phosphorus chloride = 138.06 g/mol

Explanation:

mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.

mass = 3.51 g

lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

From graham's equation of effusion,

$\frac{Rc}{Rp}$ = $\sqrt{\frac{Mp\\}{Mc} }$

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

$\frac{3.51}{1.98}$ = $\sqrt{\frac{Mp\\}{44.01} }$

1.77 = $\frac{\sqrt{Mp} }{6.64}$

11.75 = $\sqrt{Mp}$

Mp = $11.75^{2}$

Mp = molar mass of the phosphorus chloride = 138.06 g/mol

5 0

3 years ago

PLEASE HELP ITS A TESTT!!

Sergio [31]

Answer:

Explanation:

6 0

3 years ago

A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

2 years ago