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3 years ago

Write the electronic configuration of calcium oxide

Chemistry

1 answer:

Nana76 [90]3 years ago

6 0

Answer:

The electron configuration for calcium is: 1s2 2s2 2p6 3s2 3p6 4s2. Since calcium is in the fourth row and the second column of the s-block on the periodic table of elements, its electron configuration ends in 4s2. Every lower orbital is filled, starting with the 1s orbital.

Explanation:hope this helps

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Do all electrodes make good canthodes and anodes? why or why not

Sedbober [7]

An electrode is an electrical conductor used to make contact with a nonmetallic part of a circuit The word was coined by William Wheel at the request of the scientist Michael Faraday from the Greek words electron, meaning amber and hods, a way.

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3 years ago

Question 9 (2 points)

KonstantinChe [14]

Answer:

0.39 moles

Explanation:

To find how many moles are in 50.0 g of CaC₂O₄ you divide the grams of the sample by the molar mass of the compound;

$\frac{50.0 g}{128.097 g/mol}$ =0.39 mol

The grams cancel out and you are left with moles!

I hope this help ^-^

7 0

2 years ago

Read 2 more answers

How many grams of O are in 635 g of Li 2 O ?

KiRa [710]

Answer:

340g

Explanation:

Lithium oxide or Li2O is an inorganic compound made of two lithiums and one oxygen molecules. Lithium molecular mass is 6.94g/mol while oxygen molecular mass is 16g/mol. The molecular mass of Li2O will be:

2* 6.94g/mol + 1*16g/mol= 29.88g/mol

Out of 1 mol lithium oxide (29.88g), there is 1 mol of oxygen(16g). Then, out of 635g lithium oxide the number of oxygen will be: 635g * (16g/29.88g)= 340g

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2 years ago

3. Which of the following phase changes requires energy to happen?

nydimaria [60]

Answer:

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2 years ago

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A sample of gas in a balloon has an initial temperature of 20. ∘C and a volume of 1.92×103 L . If the temperature changes to 68

morpeh [17]

Answer:

$\boxed{2.23 \times 10^{3} \text{ L}}$

Explanation:

The pressure is constant, so we can use Charles' Law.

$\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}$

Data:

V₁ = 1.92 × 10³ L; T₁ = 20 °C

V₂ = ?; T₂ = 68 °C

Calculations:

(a) Convert temperatures to kelvins

T₁ = (20 + 273.15) K = 293.15 K

T₂ = (68 + 273.15) K = 341.15 K

(b) Calculate the volume

$\dfrac{ 1.92 \times 10^{3}}{293.15} = \dfrac{ V_{2}}{341.15}\\\\6.550 = \dfrac{ V_{2}}{341.15}\\\\V_{2} = 6.550 \times 341.15 = 2.23 \times 10^{3} \text{ L}$

The new volume of the gas is $\boxed{2.23 \times 10^{3} \text{ L}}$ .

6 0

3 years ago

Write the electronic configuration of calcium oxide ​

Write the electronic configuration of calcium oxide