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2 years ago

Magnesium and calcium are in the same group on the periodic table.

Chemistry

1 answer:

SOVA2 [1]2 years ago

8 0

Answer: They both have two valence electrons.

Explanation:

Atoms with the same number of valence electrons normally have similar properties.

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A intramuscular medication is given at 5.00mg/kg of body weight. What is the dose in grams for a 180-lb patient?

Alex17521 [72]

0.408 gram for 180-lb patient

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3 years ago

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Are there more metals or nonmetals on the periodic table?

bogdanovich [222]

Answer:yes

Explanation:

I think so

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3 years ago

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The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of

tresset_1 [31]

Answer:

The total number of moles of gas in the mixture is 0.16939.

1.25550 grams of methane gas and 3.18848 grams of oxygen gas.

Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n = $n_1+n_2$

Using an ideal gas equation :

$PV=nRT$

$P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K$

$P=2.509 n$

Partial pressure of the methane= $p_1=0.175 atm$

Moles of the methane= $n_1$

Partial pressure of the oxygen gas= $p_2=0.250 atm$

Moles of the methane= $n_2$

Mole fraction of the methane= $\chi_1=\frac{n_1}{n_1+n_2}$

Mole fraction of the oxygen gas= $\chi_2=\frac{n_2}{n_1+n_2}$

$p_1=p\times \chi_1$ (Dalton's law)

$0.175 atm=P\times \frac{n_1}{n_1+n_2}$

$0.175 atm=(2.509 n)\times \frac{n_1}{n}$

$n_1=0.06975 mol$

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

$p_2=p\times \chi_2$ (Dalton's law)

$0.250 atm=P\times \frac{n_2}{n_1+n_2}$

$0.250 atm=(2.509 n)\times \frac{n_2}{n}$

$n_2=0.09964 mol$

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n = $n_1+n_2$

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]

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3 years ago

gas has an initial volume of 2.4 L at a pressure of 1.5 atm and a temperature of 273 K. The pressure of the gas increases to 4.5

QveST [7]

We are given with two set of conditions. To calculate the final volume, we get first the number of moles under the first condition. Using PV=nRT, the number of moles is equal to 0.16 moles. We substitute this together with the other conditions in PV=nRT, the final volume is 0.92 liters.

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3 years ago

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2al2o3 yields to 4al + 3o2 .. how many miles of oxygen are produced from the decomposition of 1.26 mil of Al2O3?

gladu [14]

Answer:

$\large \boxed{\text{2.52 mol Al}}$

Explanation:

2Al₂O₃ ⟶ 4Al +3O₂

n/mol: 1.26

The molar ratio is 4 mol Al:2 mol Al₂O₃.

$\text{Moles of Al} = \text{1.26 mol Al$_{2}$O}_{3} \times \dfrac{\text{4 mol Al}}{\text{2 mol Al$_{2}$O}_{3}}= \textbf{2.52 mol Al}\\\\\text{The reaction produces $\large \boxed{\textbf{2.52 mol Al}}$}$

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3 years ago