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nadya68 [22]
3 years ago
8

How many different elements are involved in the reaction shown? Record the answer in the grid. Plz help :))

Chemistry
1 answer:
Damm [24]3 years ago
6 0

Answer:

It would be to the fourth power

Explanation:

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What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza
puteri [66]

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

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3 years ago
Which natural occurrence is shown ? Check all that apply.
viktelen [127]

Answer:

2. flooding

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What is the volume of HCl gas required to react with excess magnesium metal to produce 6.82 L of hydrogen gas at 2.19 atm and 35
prohojiy [21]

Answer:

Explanation:

2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)

Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law

P . V = n . R .T

2.19 atm . 6.82L = n . 0.082 . 308K

(2.19 atm . 6.82L) / (0.082 . 308K) = n

0.591 mol = n

1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole

0.591 .2 = 1.182 mole of acid.

Molar mass of HCl = 36.45 g/m

1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g

Density HCl = HCl mass / HCl volume

0,118 g/mL = 43.1 g / HCl volume

43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)

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3 years ago
All of the following will be damaging to the environment except
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Adding fertilizer to stimulate the aquatic plant growth, and putting research and money into renewable energy
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