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Anna71 [15]
3 years ago
6

Name the 3 seperate metals group

Chemistry
1 answer:
umka21 [38]3 years ago
8 0

Akali Metals

Akali-Earth Metals

and Other Metals

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which number should be placed before F2 on the reactants side equation to make equation balanced? Xe + ___F2 > XeF4
Ivahew [28]

There are four F atoms on the products side.

Since two more F atoms are required on the reactant side, you multiply the number of F2 molecules by two.

So 2 should be placed in front of F2

3 0
3 years ago
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Identify the number with the correct number of significant figures for each step of the calculation and for the final answer. No
lara [203]

Answer: The answer is 167

Explanation: This is because that was right on edg. so yea heart this tho plsss

7 0
3 years ago
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The ksp of pbi2 is 1.4 x 10-8. what is the molar solubility of lead(ii iodide in a solution of 0.400 m sodium iodide?
tino4ka555 [31]
The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:

PbI2 = Pb^2+ + 2I-

We solve as follows:

Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
</span><span>1.4 x 10-8 = x(2x)^2
</span><span>1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M

The molar solubility would be </span>1.5x10^-3 M.
8 0
3 years ago
What is the mass in milligrams of 4.30 moles of sodium? use significant figures?
alexandr1967 [171]
Answer is: mass <span>of 4,30 moles of sodium</span> is 98800 mg.
n(Na) = 4,30 mol.
m(Na) = ?
m(Na) = n(Na) · M(Na).
m(Na) = 4,30 mol · 23 g/mol.
m(Na) = 98,90 g.
m(Na) = 98,90 g · 1000 mg/1g.
m(Na) = 98900 mg.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
5 0
3 years ago
A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

3 0
3 years ago
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