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kherson [118]
2 years ago
5

Part B

Chemistry
1 answer:
Y_Kistochka [10]2 years ago
4 0

Answer:

1. conduction

2. Convection

3. Radiation

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What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
ValentinkaMS [17]

Answer:

5.90

Explanation:

Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol

Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol

CH3COO- + HCl => CH3COOH + Cl-

Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol

Moles of CH3COOH formed = moles of HCl added = 0.0005 mol

pH = pKa + log([CH3COO-]/[CH3COOH])

= -log Ka + log(moles of CH3COO-/moles of CH3COOH)

= -log(1.78 x 10^(-5)) + log(0.007/0.0005)

= 5.90

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3 years ago
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3 years ago
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Suppose three neutrons are released when an atom in a sample of fissionable nuclei undergoes fission. Each of these neutrons has
Eddi Din [679]

Answer:

13 nuclei

Explanation:

From the question, the fission of one nucleus produces three neutrons which causes  more nuclei to undergo fission.

This implies that, after the first fission, three neutrons cause three nuclei to undergo fission. The three nuclei that underwent fission produces nine neutrons which causes the fission of nine nuclei.

All together we the number of nuclei that underwent fission as;

1 + 3 + 9 = 13 nuclei.

3 0
2 years ago
How many moles of carbon dioxide should be produced from the decomposition of 0.100 g of sodium bicarbonate?
Gre4nikov [31]
The chemical symbol for sodium bicarbonate is NaHCO3. Its molar mass is 84 g/mol. In each of one mol of sodium bicarbonate their is one mole of carbon dioxide with the molar mass of 44 grams per mol. We determine the number of moles in 0.10 g of sodium bicarbonate. 
                         n sodium bicarbonate = (0.10 g) / (84 grams / mol)
                                                            = 1.19 x 10-3 moles sodium bicarbonate
Therefore, there are also 1.19 x 10-3 moles of carbon dioxide. 
3 0
3 years ago
Ima 5% glucose solution, how many grams of glucose would be present per 100mL
Sedaia [141]

Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>

<u />

<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>

<em><u /></em>

The solute can be <em>crystalline or non-crystalline in nature.</em>

<em></em>

The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

  • This question is based on a Percentage Weight-in-volume. The formula states that:

a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>

Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution

Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.

To learn more, visit the link below:

brainly.com/question/8482854

6 0
1 year ago
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