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alexdok [17]
3 years ago
7

You and your sister both have a small camping tent set up in the backyard. You each have two strings of lights to hang inside th

e tent. You power your lights with a 9-volt battery and your sister uses an extension cord and powers her lights from an electrical outlet on the deck.
What is the BEST conclusion the you can make about this situation?
A
Your lights shine the brightest because your battery produces less power.

B
Your lights shine the brightest because your battery produces more power.

C
Your sister's lights shine the brightest because the electrical outlet has less power.

D
Your sister's lights shine the brightest because the electrical outlet has more power.
Chemistry
1 answer:
hoa [83]3 years ago
6 0

Answer:

D

Your sister's lights shine the brightest because the electrical outlet has more power.

Explanation:

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Why is graphite used for electrodes?
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Which of the following combinations can be used to prepare a buffer at ph 1​
xeze [42]

A buffer is a solution which resists changes to it pH when a small quantity of acid or base is added to it.

<h3>What is a buffer?</h3>

A buffer is a solution which resists changes to it pH when a small quantity of acid or base is added to it.

Buffers are made from solutions of weak acids and their salts or weak bases and their salts.

<h3>What is pH?</h3>

The pH of a solution is the negative logarithm to base 10 of yhe hydrogen ion concentration of the solution.

Solutions of low pH have high concentration of hydrogen ions while solutions of high pH have high hydrogen ion concentration.

Learn more about about buffers and pH at: brainly.com/question/22390063

7 0
3 years ago
How many moles of water were lost if the amount of water lost was 0.294 grams?
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7 0
3 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
3 years ago
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