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babymother [125]
2 years ago
9

True or False: Quantitative data is about a description or observation and qualitative data is about measurements and numbers. P

lease help in under 20 min,
Chemistry
2 answers:
kati45 [8]2 years ago
4 0
False, quantitative is numerical date, qualitative is about descriptive or observation data.
Mice21 [21]2 years ago
3 0

Answer:

False

Explanation:

Quantitative data is measured using numbers and Qualitative data is descriptive

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A cube has an edge length of 9 cm .
Angelina_Jolie [31]

Answer:

9^3 i think so like 279

Explanation:

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5 0
3 years ago
1 gallon is how many mm
leva [86]

Answer:

0.133681 (cubic foot)

Explanation:

6 0
2 years ago
What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?
Trava [24]
Answer is: pH value of weak is 3.35.
Chemical reaction (dissociation): HA(aq) → H⁺(aq) + A⁻(aq).
c(HA) = 0.0055 M.
α = 8.2% ÷ 100% = 0.082.
[H⁺] = c(HA) · α.
[H⁺] = 0.0055 M · 0.082.
[H⁺] = 0.000451 M.
pH = -log[H⁺].
pH = -log(0.000451 M).
pH = 3.35.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity <span>an aqueous solution.</span>
3 0
3 years ago
All of the following methods would increase the solubility of a solid solute, except for
cricket20 [7]
May i please have a(n) answer choices please because it would be a lot better if it was like that and then ill answer it
8 0
3 years ago
I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine so
sergeinik [125]

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

<em>Moles C2H5NH2:</em>

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

<em>Moles HNO3:</em>

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

6 0
2 years ago
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