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3 years ago

5

Which of the following elements is least likely to form a cation that will then form an ionic bond with an ion of a Group 7A ele

ment?
K
Rn
F
Fr

1 answer:

dexar [7]3 years ago

8 0

F because it NEVER forms any cations in chemical reactions

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Which of the following best captures the point of paragraph 3

givi [52]

Answer:

Is there a picture or smth?

Explanation:

6 0

2 years ago

Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

creativ13 [48]

The question is incomplete, here is the complete question:

A solution contains 0.115 mol $H_2O$ and an unknown number of moles of sodium chloride. The vapor pressure of the solution at 30°C is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

<u>Answer:</u> The mass of sodium chloride in the solution is 0.714 grams

<u>Explanation:</u>

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (water) = 31.8 torr

$p^s$ = vapor pressure of the solution = 25.7 torr

i = Van't Hoff factor = 2

$\chi_{\text{solute}}$ = mole fraction of solute (sodium chloride) = ?

Putting values in above equation, we get:

$\frac{31.8-25.7}{31.8}=2\times \chi_{NaCl}\\\\\chi_{NaCl}=0.0959$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{NaCl}}=\frac{n_{\text{NaCl}}}{n_{\text{NaCl}}+n_{\text{water}}}$

We are given:

Moles of water = 0.115 moles

$0.0959=\frac{n_{\text{NaCl}}}{n_{\text{NaCl}}+0.115}\\\\n_{\text{NaCl}}=0.0122mol$

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of NaCl = 0.0122 moles

Molar mass of NaCl = 58.5 g/mol

Putting values in above equation, we get:

$0.0122mol=\frac{\text{Mass of NaCl}}{58.5g/mol}\\\\\text{Mass of NaCl}(0.0122mol\times 58.5g/mol)=0.714g$

Hence, the mass of sodium chloride in the solution is 0.714 grams

5 0

3 years ago

Which of the following systems has potential energy only?

bagirrra123 [75]

Answer D

Explanation:

6 0

2 years ago

How many liters of oxygen gas, at standard

Karo-lina-s [1.5K]

Answer:

Explanation:

For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

3 years ago