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Semmy [17]
3 years ago
11

I need help ASAP pls

Chemistry
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

A  should be the swer n

Explanation:

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Simple ways to reduce heat loss include fitting carpets, curtains and draught excluders. It is even possible to fit reflective foil in the walls or on them. Heat loss through windows can be reduced by using double glazing. These special windows have air or a vacuum between two panes of glass.
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3 years ago
g Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.
jarptica [38.1K]

Answer:

Following are the responses to the given choices:

Explanation:

For point a:

Using the acid and base which are strong so,

moles of H^+ (fromHNO_3)

= 24.9\ mL \times 0.100\ M \\\\= \frac{24.9}{1000\ L} \times 0.100\  M \\\\= 2.49 \times 10^{-3} \ mol

moles of OH^{-} (from KOH)

= 25.0\ mL \times 0.100\ M \\\\= \frac{25.0}{1000 \ L} \times 0.100 \ M \\\\\= 2.50 \times  10^{-3}\  mol  

1\ mol H^{+} \ neutralizes\  1\ mol\  of\  OH^{-}

So,  (2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol) i.e. 1 \times 10^{-5} mol of OH^- in excess in total volume (24.9+25.0) \ mL = 49.9 \ mL i.e. concentration of OH^- = 2 \times 10^{-4}\ M

p[OH^{-}] = -\log [OH^{-}] = -\log [2 \times 10^{-4}\ mol] = 3.70

Since, pH + pOH = 14,

so,

\to pH = 14- pOH = 14- 3.70 = 10.30  

For point b:

moles of OH^- = from point a = 2.50 \times 10^{-3} \ mol

moles of H^+(fromHNO_3):

= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol

1 mol H^+ neutralizes 1 mol of OH^-

So, (2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol) i.e. 1 \times 10^{-5} \ mol \ of\  H^+ in excess in the total volume of (25.1+25.0) \ mL = 50.1\ mL i.e. concentration ofH^+ = 2 \times 10^{-4}\  M

Hence, pH = -\log [H^+] = -\log[2 \times 10^{-4}] = 3.70

6 0
3 years ago
What volume of a 0.155 M potassium hydroxide solution is required to neutralize 25.7 mL of a 0.388 M hydrobromic acid solution
vekshin1

Answer: Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HBr solution = 0.338 M

V_1 = volume of HBr solution = 25.7 ml

M_2 = molarity of KOH solution = 0.155 M

V_2 = volume of KOH solution = ?

n_1 = valency of HBr = 1

n_2 = valency of KOH = 1

1\times 0.338\times 25.7=1\times 0.155\times V_2

V_2=56.0ml

Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

8 0
3 years ago
How to write a chemical formula ?​
Kaylis [27]

Answer:

To write a chemical equation, the reactants should be written on the left, and the products should be written on the right and the coefficients should be written next to the symbols of entities to indicate the number of moles of a substance produced used in the chemical reaction.

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204.0920-It have 7 significant digits
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