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Gelneren [198K]
2 years ago
9

Lead(II) chromate, PbCrO4, is a yellow paint pigment (called chrome yellow) prepared by a precipitation reaction. In a preparati

on, 52.5 g of lead(II) chromate is obtained as a precipitate. How many moles of PbCrO4 is this
Chemistry
1 answer:
Roman55 [17]2 years ago
8 0

Answer:

n = 0.1624 moles

Explanation:

n=\frac{m}{M} \\n=\frac{52.5}{323.1937} \\n= 0.1624 mol

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Answer:

reactivaty

Explanation:

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5 0
3 years ago
The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

4 0
3 years ago
Lanthanium found in the......... period
Gnesinka [82]
I'm preatty sure it is in the periodic table xxx

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3 years ago
Which of the following could be considered a scientific statement?
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7 0
3 years ago
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how many grams of calcium oxide are producer from the decomposition of 125 grams of calcium carbonate
Dmitrij [34]

70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.

MM = 100.09 56.08

CaCO3 → CaO + CO2

Mass 125 g

a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)

= 1.249 mol CaCO3

b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)

= 1.249 mol CaO

c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g

7 0
3 years ago
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