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Ber [7]
3 years ago
14

How many pennies are in $2,020.20?

Chemistry
2 answers:
Nata [24]3 years ago
5 0

Answer:

202,020 pennies

Explanation:

The amount of pennies in 1 dollar is 100.

Basically, you just multiply the whole dollar amount (2020) by 100 and then add 20 to find the amount of pennies.

(2020)(100)+20= 202,000+20= 202,020 pennies.

Goshia [24]3 years ago
5 0

Answer:

202,040 pennies

Explanation:

$1 is equal to 100 so just multiply 100 by 2,020 and then you can add the extra 20 cents

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It would be Xe because all of the other ones aren't filled up in the correct order which only happens during an excited state. 
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3 years ago
CaCl2 (picture) i don’t know how to do the problem
Vadim26 [7]

Answer:

Ca: 1

Cl: 2

Explanation:

Since there is no subscript but there is the element, its assumed that there is at least 1, otherwise it would'nt be there

8 0
4 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

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3 years ago
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Number 4 should be physical change since scientists are breaking the water into 2 elements, O₂ and H₂

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