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Semmy [17]
3 years ago
10

If 0.2 M Ag+, 2 M Ba2+, and 2 M Ca2+ are added to a Na2SO4 solution, which will precipitate first? (Ksp of Ag2SO4 = 1.4 x 10-5,

Ksp of BaSO4 = 1.1 x 10-10, and Ksp of CaSO4 = 9.1 x 10-6)
A.) BaSO4
B.) Ag2SO4
C.) BaSO4 and CaSO4
D.) Ag2SO4 and CaSO4
Chemistry
1 answer:
Sedbober [7]3 years ago
7 0
A.

This is because BaSO₄ has the lowest solubility product out of all the substances.
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3 years ago
Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40
devlian [24]

Answer:

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

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Initial temperature = 20°C

Final temperature = 40°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

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ΔT =  40°C -20°C

ΔT =  20°C

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48 J =240 g.°C×c

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6 0
3 years ago
I need help on homework
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3 0
4 years ago
Diamond has a density of 3.513 g/cm³. The mass of diamonds is often measured in "carats," where 1 carat equals 0.200
Gennadij [26K]

The volume of a 1.86-carat diamond in cubic centimeters is 0.106 cm³

Given,

The density of a diamond is  3.513 g/cm³.

We have to find out the volume of a 1.86-carat diamond in cubic centimeters.

Convert the units of the diamond from carat to grams, we have:

(1.86 carats) x (0.200 g / 1 carat) = 0.372 g

The volume of the diamond is obtained by dividing the mass by the density, therefore using the formula, we get

                            v = m / d

                            v = 0.372 g / (3.51 g/cm³) = 0.1059 cm³

                       or, v = 0.106 cm³ (approx)

Therefore, the volume of a 1.86-carat diamond is approximately 0.106 cm³.

To learn more about the volume, visit: brainly.com/question/1578538

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4 0
2 years ago
Calculate the de broglie wavelength of a subatomic particle that is moving at 351 km/s if its mass is 9.11 à 10â31 kg. λ = hmuh
gregori [183]

The de Broglie wavelength of a subatomic particle is 2.09 nm.

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calculation is given in the image below

de Broglie wavelength λ = h/mv

                                          = (6.626 * 10^-34)/9.1 * 10^-31 *351 *10^3

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6 0
2 years ago
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