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2 years ago

3 3The number of electrons in Ne and Mg2+ is same. Give reason,

Chemistry

1 answer:

motikmotik2 years ago

6 0

Explanation:

mg 2+ losses 2 electron due to the charge on it making it change from 12 to 10.and ne is also 10

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What is the gcf of 16 and 72? a. 2 b. 4 c. 16 d. 8

nadya68 [22]

Your answer is D. 8

16 = 2^4
72 = 2^3*3^2

So you'll choose 2^3 = 8

7 0

3 years ago

Read 2 more answers

126785033% Divided by 348675948849 = what does it equal

meriva

Answer:

0.00000363618

could be wrong.

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5 0

3 years ago

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

kenny6666 [7]

Answer:

Lithium oxide, Li₂O.

Explanation:

Hello!

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

$M_2O+H_2\rightarrow 2M+H_2O$

Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

$5.00gM_2O*\frac{1molM_2O}{(2*X+16)gM_2O}*\frac{2molM}{1molM_2O}*\frac{XgM}{1molM}=2.32gM$

Thus, we perform the operations in order to obtain:

$\frac{10X}{(2X+16)}=2.32$

So we solve for x as shown below:

$10X=2.32(2X+16)\\\\10X=4.64X+37.12\\\\X=\frac{37.12}{10-4.64}\\\\X= 6.93g/mol$

Whose molar mass corresponds to lithium, and therefore, the metallic oxide is lithium oxide, Li₂O.

Best regards!

8 0

3 years ago

Write word equation followed by chemical equation for the following reactions:a) iron with copper (ll) sulphate b) copper with s

8_murik_8 [283]

Answer:

Iron + copper(II) sulphate → iron sulphate + copper

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

copper + silver nitrate → copper nitrate + silver

Cu (s) + 2AgNO3 (aq) → Cu(NO3)2 (aq) + 2Ag (s)

3 0

3 years ago

a 5L container contains 3 moles of helium and 4 moles of hydrogen at a pressure of 9 atms maintaining a constant T and additiona

Stells [14]

Answer:

7.71 atm

Explanation:

Given the following data:

$V = 5 L$

$n_{He} = 3 mol$

$n_{H_2} = 4 mol$

$p_1 = 9 atm$

$T = const$

According to the ideal gas law, we know that the product between pressure and volume of a gas is equal to the product between moles, the ideal gas law constant and the absolute temperature:

$pV = nRT$

Since the temperature and the ideal gas constant are constants, as well as the fixed container volume of 5 L, we may rearrange the equation as:

$\frac{p}{n}=\frac{RT}{V}=const$

This means for two conditions, we'd obtain:

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

Given:

$p_1 = 9 atm$

$n_1 = n_{initial total} = n_{He} + n_{H_2} = 3 mol + 4 mol = 7 mol$

$n_2 = n_{final total} = n_{He} + n_{H_2} = 3 mol + 4 mol + 2 mol = 9 mol$

Solve for the final pressure:

$p_2 = p_1\cdot \frac{n_2}{n_1}$

Now, according to the Dalton's law of partial pressures, the partial pressure is equal to the total pressure multiplied by the mole fraction of a component:

$p_{H_2}=\chi_{H_2}p_2$