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il63 [147K]
3 years ago
15

It takes 208.4 kJ of energy to remove 1 mole of electrons from 1 mole of atoms on the surface of rubidium metal. How much energy

does it take to remove a single electron from an atom on the surface of solid rubidium?
Chemistry
1 answer:
Darya [45]3 years ago
4 0

Answer:

3.46*10⁻²² kJ

Explanation:

By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:

6.02*10²³ electrons ---------------- 208.4 kJ

1 electron                  ---------------- x

By a simple direct three rule:

6.02*10²³ x = 208.4

x = 3.46*10⁻²² kJ

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What type of reaction is shown below c10H8+12O2 10co2+4H2O
MakcuM [25]

Answer:

Explanation:

Combustion.  Have fun with that.

7 0
3 years ago
T
svet-max [94.6K]

Explanation:

(1) The nucleus is positive and the electron cloud is positive.

(2) The nucleus is positive and the electron cloud is negative.

(3) The nucleus is negative and the electron cloud is positive.

(4) The nucleus is negative and the electron cloud is negative

4 0
3 years ago
A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha
Ainat [17]

We have that the molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol

Therefore

The molecular weight (3sf) of the compound (g/mol)

m=44.15g/mol

For more information on this visit

brainly.com/question/17756498

4 0
2 years ago
5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?
Aleksandr [31]

Answer:

V_2=9.20L

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

\frac{V_2}{n_2}= \frac{V_1}{n_1}

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L

Best regards!

3 0
3 years ago
Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change
USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation  follows:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]

We are given:

\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

5 0
3 years ago
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