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Lyrx [107]
3 years ago
7

At the equivalence point for a weak acid-strong base titration an equal number of moles of OH- and H+ have reacted, producing a

solution of water and salt. What affects the pH at the equivalence point for a weak-acid/strong-base titration? The basicity of the salt anion The acidity of the salt cation The auto-ionization of water None of these
Chemistry
1 answer:
aev [14]3 years ago
6 0

Answer: The basicity of the salt anion

Explanation:

All the weak acid is neutralized at the equivalence point and transferred to its conjugate base (number of moles of H+ = increased number of moles of OH–). Therefore the solution produced is weakly alkaline and the pH of the equivalence point will be greater than 7.

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The table shows the amount of radioactive element remaining in a sample over a period of time.
Lunna [17]

Answer:

1. The half-life is 22 years.

2. 132 years

Explanation:

1. Determination of the the half-life.

The half-life of an element is the time taken for half the element to decay.

From the table given above, the original amount of the element 45 g. If we divide 45 by 2, we'll have 22.5 g as half the original amount of element.

Now, the time taken to obtain 22.5 g as shown from the table is 22 years.

Thus, the half-life the element is 22 years.

2. Determination of the time.

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Time (t) =?

Next, we shall the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

4.8125 = 1/2ⁿ × 308

Cross multiply

4.8125 × 2ⁿ = 308

Divide both side by 4.8125

2ⁿ = 308 / 4.8125

2ⁿ = 64

Express 64 in index form with 2 as the base.

2ⁿ = 2⁶

n = 6

Thus, 6 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Number of half-lives (n) = 6

Half-life (t½) = 22 years

Time (t) =?

n = t / t½

6= t / 22 years

Cross multiply

t = 6 ×22

t = 132 years.

Thus, the time taken is 132 years.

6 0
3 years ago
during the processes of erosion and deposition sediments that are the ____ in size will be carried the greatest distances before
Harlamova29_29 [7]
<span>Smallest sediments will go the farthest. Usually sand or silt size grains.</span>
8 0
3 years ago
What is the mass of of 3.20 X 10^23 particles of Co2?​
vitfil [10]

Answer:

Mass = 23.36 g

Explanation:

Given data:

Number of particles of CO₂ = 3.20 ×10²³

Mass of  CO₂ = ?

Solution:

1 mole contain 6.022×`10²³ particles,

3.20 ×10²³ particles  × 1 mol / 6.022×`10²³ particles

0.531 mol

Mass of CO₂:

Mass = number of moles × molar mass

Molar mass = 44 g/mol

Mass =  0.531 mol× 44 g/mol

Mass = 23.36 g

6 0
3 years ago
According to the Bronsted Lowery definition, a base is a a substance that increases the hydroxide ion concentration in water b a
Scrat [10]

Answer:

B. a substance that can accept a proton from an acid

Explanation:

Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions).

The acid is the donor of the hydrogen ions while the bass is the acceptor of the hydrogen ions.

Back to the options, the correct option is B. Which is a substance that can accept a proton from an acid. All other options are incorrect.

7 0
3 years ago
Phosphorus is obtained primarily from ores containing calcium phosphate.
ladessa [460]

Answer:

33.7 kg

Explanation:

Let's consider calcium phosphate Ca₃(PO₄)₂.

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol and the molar mass of P is 30.97 g/mol. In 1 mole of Ca₃(PO₄)₂ (310.18 g) there are 2 × 30.97 g = 61.94 g of P. The mass of Ca₃(PO₄)₂ that contains 3.57 kg (3.57 × 10³ g) of P is:

3.57 × 10³ g × (310.18 g Ca₃(PO₄)₂/61.94 g P) = 1.79 × 10⁴ g Ca₃(PO₄)₂

A particular ore contains 53.1% calcium phosphate. The mass of the ore that contains 1.79 × 10⁴ g of Ca₃(PO₄)₂ is:

1.79 × 10⁴ g Ca₃(PO₄)₂ × (100 g Ore/ 53.1 g Ca₃(PO₄)₂) = 3.37 × 10⁴ g Ore = 33.7 kg Ore

8 0
3 years ago
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