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2 years ago

In what orbital would to valence electrons of strontium be found? a.) 5s b.) 4s c.) none d.) 5p

Chemistry

1 answer:

KonstantinChe [14]2 years ago

4 0

Answer:

The correct answer is 5s

Explanation:

Strontium atomic number = 38

Electronic configuration is $$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 5s^{2}$

As $5s$ orbital is valence orbital in strontium, the outer electrons exists in that orbital only.

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Hich of the following statements is true about what happens during a chemical reaction?

postnew [5]

I believe it is <span>d. the bonds of both the reactants and the products are formed.</span>

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2 years ago

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0

2 years ago

Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?

TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

$2Al+6HBr\rightarrow 2AlBr_3+3H_2$

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

$8molAl(\frac{6molHBr}{2molAl})$

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

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2 years ago

If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced

marissa [1.9K]

The amount in grams of Al₂O₃ produced is approximately 6.80 g.

Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:

AL + O₂ → Al₂O₃

Let's balance it

4AL + 3O₂ → 2Al₂O₃

4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,

Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,

Atomic mass of AL = 27 g

Molar mass of Al₂O₃ = 101.96 g/mol

4(27 g) of AL gives 2(101.96 g) of Al₂O₃

3.6 g of AL will give ?

cross multiply

mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797

mass of Al₂O₃ produced = 6.80 g.

8 0

2 years ago

Which information in the periodic table indicates the number of protons in an atom? A. the position of the element in its c

Leni [432]

Hi there!

D is true

the number of protons is same as atomic number in an atom

good luck!

3 0

3 years ago

Read 2 more answers

In what orbital would to valence electrons of strontium be found? a.) 5s b.) 4s c.) none d.) 5p​

In what orbital would to valence electrons of strontium be found? a.) 5s b.) 4s c.) none d.) 5p