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Vika [28.1K]
3 years ago
8

What is the electron configuration for a calcium atom? 1s2, 2s2, 2p6, 3s2, 3p6, 4s2 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 2s2, 2s2, 2p6,

3s2, 3p6, 4s1 1s1, 2s2, 2p6, 3s2, 3p6, 4s1
Chemistry
2 answers:
Zarrin [17]3 years ago
8 0
1s2,2s2, 2p6, 3s2,3p6,4s2
snow_tiger [21]3 years ago
4 0

Its 1s2 2s2 2p6 3s2 3p6 4s2

Its in Group 2 of the Periodic Table

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The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st
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Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.

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After comparing DNA, a doctor finds that two different samples are the same. The two individuals from whom the samples came are
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B identical twins is the answer I believe
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What is use in measuring
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Answer:

a ruler or measuring cups are ways of measuring

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6 0
2 years ago
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What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
BigorU [14]
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.
6 0
3 years ago
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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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