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2 years ago

Gu

Chemistry

1 answer:

Andrei [34K]2 years ago

7 0

According of Dalton's law of Partial pressure, the total pressure of a mixture of gases is the sum of the partial pressures of the individual vases in the mixture.

Hence;

The for hydrogen collected over water, we have a mixture of hydrogen gas and water vapour.

Total pressure = pressure of hydrogen gas + vapour pressure of water

Pressure of hydrogen gas = Total pressure - vapour pressure of water

Pressure of hydrogen gas = 636 mmHg - 28.3 mmHg

Pressure of hydrogen gas = 607.7 mmHg

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Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the

Alecsey [184]

Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

$\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

102.5 = $(\frac{1}{2})x + 498 - 486$

102.5 - 12 = $\frac{x}{2}$

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

$\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

= 339.5 - 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0

3 years ago

If a radio station transmits on AM 610, how many hertz (Hz) is the frequency of the wave? (Remember that kHz = kilohertz.)

elixir [45]

Answer:

610,000

Explanation:

8 0

3 years ago

Help asap on a time crunch

marusya05 [52]

Answer:

only sodium chloride compounds MOST LIKELY products of the reaction between sodium metal and chlorine gas.

Correct answer - C

Explanation:

A solid sodium metal reacts with gaseous form of chlorine gas to form a solid sodium chloride.

This formation of sodium chloride arises due to transfer of one valence electron from sodium to chlorine atom and the both atoms get stable octet electronic configuration.

As a result, an ionic compound solid sodium chloride is formed.

The chemical reaction of formation sodium chloride is as follows.

$2Na(s)+Cl_{2}(g) \rightarrow 2NaCl(s)$

Therefore, only sodium chloride compounds MOST LIKELY products of the reaction between sodium metal and chlorine gas.

7 0

3 years ago

Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

(0.75M) x (0.094L) = C2

(0.329L)

C2 = 0.214 M (Rounded)

When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.

7 0

2 years ago

Given 7.40 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

Dafna11 [192]

On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....

4 0

3 years ago