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2 years ago

Pls help

Chemistry

1 answer:

Irina-Kira [14]2 years ago

3 0

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0

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4. C. 6

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3 years ago

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The density of Nitrogen (N2) gas in a 4.32 L container at our

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Answer:

δ N2(g) = 1.1825 g/L

Explanation:

δ ≡ m/v
Mw N2(g) = 28.0134 g/mol

ideal gas:

PV = RTn

∴ P = (837 torr)×( atm/760 torr) = 1.1013 atm

∴ T = 45.0 °C + 273.15 = 318.15 K

∴ R = 0.082 atm.L/K.mol

⇒ n/V = P/R.T

⇒ n/V = (1.1013 atm) / ((0.082 atm.L/K.mol)(318.15 k))

⇒ n/V = 0.0422 mol/L

⇒ δ N2(g) = (0.042 mol/L)×(28.0134 g/mol) = 1.1825 g/L

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3 years ago

What is catalyst? Write its types.

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Answer:

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3 years ago

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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3 years ago

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

<em>c) 0.1 M NaClO or</em>

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water

I hope it helps!

7 0

3 years ago