Chemical equation to show cdso4 dissolves in water

Chemistry

1 answer:

olga_2 [115]3 years ago

5 0

no se bb si quieres salimos tu sabes

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How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex

Brut [27]

When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M
so by substitution we can get [Ca2+]
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
= 0.00167 mol

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3 years ago

Which describes the products created from fission reactions? Question 3 options:

Ipatiy [6.2K]

<u>Answer:</u> The correct answer is option A.

<u>Explanation:</u>

Nuclear fission reactions are a type of nuclear reactions in which larger nuclei breaks apart into two or more smaller fragment releasing alpha, gamma of beta particles.

There are 3 types of particles that can be released during this process:

1. Alpha particles: These particles are released when a nuclei undergoes alpha-decay process.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

2. Beta particles: These particles are released when a nuclei undergoes beta-minus decay process.

$_Z^A\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^0\beta$

3. Gamma radiations: these radiations are released when an unstable nuclei gives off excess energy by a process of spontaneous electromagnetic process.

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

Hence, any of these particles can be released during the process of fission reaction with smaller atoms.

Therefore, the correct answer is option A.

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3 years ago

A 0.5376 g sample of an unknown compound is found to contain 0.3044 g of carbonate. Could this compound be calcium carbonate?

shepuryov [24]

Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams

One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams

Based on the above calculations, the sample is not CaCO3

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