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Oksana_A [137]
3 years ago
14

Why do they put flexible connections in the road on a bridge

Chemistry
1 answer:
Novay_Z [31]3 years ago
6 0
This is because the bridge is completely surrounded by cold air in the winter and by hot air in the summer. It is not insulated by the ground beneath it.
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The acidity and basicity of a solution can be described by its ph. a solution will be more acidic if
balandron [24]
If the ph is a low value, below 7, it would be acidic.
8 0
3 years ago
What percent of the human body is water
Harlamova29_29 [7]

Answer:

Up to 60% of the human adult body is water.

4 0
3 years ago
Read 2 more answers
Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B Mass ratio A/B =
LUCKY_DIMON [66]

Answer:

mass ratio of A/B is 2:1

Explanation:

Since the mass of box A = 10g

                mass of box B = 5g

   Mass of box C = mass of box A + mass of box

A ratio compares two quantities. To find the ratio of the two boxes:

  Ratio of A to B = \frac{mass of A}{mass of B}

  Ratio of A to B = \frac{10grams}{5grams} = 2

The mass ratio is 2:1 i.e box A has twice the mass of B

8 0
3 years ago
What is the correct formula for the
erastova [34]

Answer:

A

Explanation:

cause its easy

3 0
3 years ago
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An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the
serg [7]

Answer:

pH = 12.43

Explanation:

<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

<em>Moles KOH:</em>

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

<em>Moles HN3:</em>

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

6 0
3 years ago
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