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Alika [10]
4 years ago
14

NEED HELP ASAP WILL GIVE BRAINLIEST. What is the element that has the atomic number of 17?

Chemistry
1 answer:
salantis [7]4 years ago
5 0

Hey buddy the answer is Chlorine ..

plzz follow me I'll help u more if u need

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The only positive ion found in H2SO4(aq) is the
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The answer is ammonium iron

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PLEASE PLEASE BRAINLIEST HELP ME PLEASE SOMEONE VARY SMART PLEASE SOMEONE THAT IS ACE PLEASE OR AN EXPERT OR SOMEONE FROME THE L
ValentinkaMS [17]

Answer:

1: A

2: C

3: D

4: D

5: B

Explanation:

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4 years ago
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An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
Anna [14]

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

5 0
3 years ago
Which of the following statements is true of an aqueous solution of sodium chloride?
Ilya [14]

Answer:

What are the statements please

Explanation:

4 0
3 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
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