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zysi [14]
2 years ago
14

Volume x Density = ?

Chemistry
1 answer:
Anastasy [175]2 years ago
7 0

Answer: Mass

Explanation:

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A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi
finlep [7]

Answer:

hypochlorite ion

Explanation:

The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:

HCl + ClO⁻ → HClO + Cl⁻.

Where more hypochlorous acid is produced.

That means, the HCl reacts with the hypochlorite ion present in solution

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3 years ago
Anyone could help me out?
Hoochie [10]

1.4 mg/dL = 0.014 g/L

Explanation:

Milligrams per deciliter to grams per liter

There is 1000 grams of mg/dL of 1 g/L

8 0
3 years ago
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What kind of sport do you have interest the most and explain why? <br>​
lidiya [134]

Answer

Gymnastics

Explanation:

for the simple fact that they do cool stuff

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3 years ago
What are prevailing winds?
Tems11 [23]

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

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7 0
3 years ago
A cylindrical glass tube of length 27.4cm and radius 4 cm is filled with gas. The empty tube has a mass of 254.3 g. The tube fil
Temka [501]

Density of the gas is 3.05 × 10⁻³ g / cm³.

<u>Explanation:</u>

Volume of the cylinder = π r² h

where r is the radius and h is the height of the height or the length of the glass tube.

Here r = 4 cm and h = 27.4 cm

Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³

We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.

Mass of the substance = 258.5 - 254.3 = 4.2 g

We have to find the density using the formula as,

$ Density = \frac{mass}{volume}

Plugin the values as,

$ Density = \frac{4.2}{1376.6}

              = 3.05 × 10⁻³ g / cm³

So the Density of the gas is 3.05 × 10⁻³ g / cm³.

3 0
3 years ago
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