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Assoli18 [71]
3 years ago
7

A chemist weighed out 70.8 g of magnesium. Calculate the number of moles of magnesium she weighed out. Round your answer to sign

ificant digits.

Chemistry
2 answers:
solniwko [45]3 years ago
8 0

Answer:

2.91 grams of Magnesium

Explanation:

I put a picture . I used the molar mass of magnesium which can be calculated from the periodic table or simply google it. :) I hope it helps - Enrique

AlekseyPX3 years ago
4 0
You first take the weighed out amount of magnesium then you divide by the actual mass of magnesium 24.305 g. 70.8/24.305 = 2.912980868134129. Then you multiply by Avagadros number 6.02x10^23. 2.9129 x 6.02x10^23 = 1.753614482617^24. Then round it to 1.75^24 moles of magnesium
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2Al + 3H2SO4 -> Al2(SO4)3 + 3H2How many grams of aluminum sulfate would be formed if 250g H2SO4 completely reacted with alumi
bazaltina [42]

Answer:

290.82g

Explanation:

The equation for the reaction is given below:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol

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Now, we can obtain the mass of aluminium sulphate formed by doing the following:

From the equation above:

294g of H2SO4 produced 342g of Al2(SO4)3.

Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3

Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.

7 0
3 years ago
The dissolution of barium hydroxide in water is an exothermic process. Which of the following statements is correct?A) The entha
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8 0
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Answer:

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3 quantitative physcial

4 quantiative chemical

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6 0
3 years ago
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

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The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

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pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

7 0
3 years ago
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