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garik1379 [7]
3 years ago
13

If there are 20.0g of KOH available, how many grams Ca(OH)2 will form?

Chemistry
1 answer:
notka56 [123]3 years ago
6 0

Answer:

can u send the qsn 4rm the start so tt I can help

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Two elements that have properties similar to those of the element potassium(K). To those of krypton(Kr)
balu736 [363]

Answer:

Calcium (Ca) and Scandinavium (Sc)

Explanation:

They are all on the same period.

8 0
3 years ago
One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng
tigry1 [53]

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length l=0.700nm=>0.7*10^9nm

Density \rho=2.32g/cm^3

Generally the equation for Volume is mathematically given by

V=l^3

V=(0.7*10^9)m^3

V=3,43*10^-{22}cm

Where

Molar mass of  (SiO2) for one formula unit

M=28+32

M=60g/mol

Therefore

Density of Si per unit length is

\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}

\rho=0.29

Molar mass of  (SiO2) for one formula unit

M=28+32

M=60g/mol

Therefore

There are 8Si atoms and 16 O atoms per unit cell

6 0
3 years ago
A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a
Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0
3 years ago
Naturally occurring boron has an atomic weight of 10.811. Its principal isotopes are 10B and 11B. Part A What is the abundance (
Darya [45]

Answer:

10B has 18.9%

11B has 81.1%

Explanation:Please see attachment for explanation

4 0
3 years ago
Please help me!!!!
miv72 [106K]

Answer:

0.8g/ml

Explanation:

d=m/v

d=20/25

d=0.8

7 0
3 years ago
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