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zhenek [66]
3 years ago
5

The atomic mass of the isotope copper-__ is 62.930 amu

Chemistry
1 answer:
yarga [219]3 years ago
8 0
The correct answer is copper 63
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The pH of a solution is 5.0. To a 10 ml of this
mylen [45]

Answer:

pH = 7

Explanation:

pH is defined as -log [H⁺]. In a solution of pH 5:

pH = -log [H⁺]

10^(-5) = [H⁺]

1x10⁻⁵M = [H⁺].

Then, this solution is diluted from 10mL to 1000mL (990mL of water + 10mL of the original solution). That means the solution is diluted:

1000mL / 10mL / 100 times.

If [H⁺] before dilution was 1x10⁻⁵M, after dilution will be:

1x10⁻⁵M / 100 = 1x10⁻⁷M

And pH:

pH = -log [H⁺]

pH = -log 1x10⁻⁷M

<h3>pH = 7</h3>
8 0
3 years ago
How many atoms are in 2 moles of phosphorus
Goryan [66]
Answer 2: 1 mole = 6.03 x 1023 particles. One mole of any element has a mass in grams that is equal to its atomic number, and has exactly 6.02 x 1023 atoms - however because the atoms of each element have different sizes and weights, then the volume that each one occupies is different.

Credits to
https://scienceline.ucsb.edu/getkey.php?key=274
4 0
3 years ago
The pH of a solution in which the concentration of H+ is 0.010M will be:
shusha [124]

B: 2

If it's right plz give brainliest lol <3

7 0
2 years ago
HNO3 and H2CO3 are examples of ?
Basile [38]
A) acids because they start with h
7 0
3 years ago
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
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