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kupik [55]
3 years ago
9

What is the ph of a solution that results from mixing 25 ml of 0.15 M hcl to 25 ml of 0.52 m nh3?

Chemistry
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

7.00

Explanation:

When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.

First, let's find the number of moles of each one of them, multiplying the concentration by the volume:

nH+ = 0.15 M * 25 mL = 3.75 mmol

nNH3 = 0.52 M * 25 mL = 13 mmol

So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.

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2) A balloon was inflated to a volume of 5.0 liters at a temperature of
TEA [102]

Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L

So, the new volume is 6.12 L.

8 0
2 years ago
How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +
aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

                                                                   = -3608 kJ

                                                                   ≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

5 0
3 years ago
Can someone help me with number 4? Please and thank you
kipiarov [429]

Answer:

1st Blank: <em>1 Co</em>

2nd Blank:<em> 2 Na2S</em>

3rd Blank:<em> 4 Na</em>

4th Blank:<em> 1 CoS2</em>

Explanation:

<em>Trust me</em>

6 0
3 years ago
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xxMikexx [17]
PH = -log10 [H+]. So anwer 2 pH
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3 years ago
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