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3 years ago

In which of the ions do all x-o bonds (x indicates the central atom) have the same length?

Chemistry

1 answer:

Radda [10]3 years ago

7 0

<span>Answer: All of the ions do all x-o bonds (x indicates the central atom) have the same length.</span>

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

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

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

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, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 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 = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

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 = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

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

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

What is FeBr3 compound name?

inessss [21]

Iron bromide. Iron bromide (FeBr3)

3 0

2 years ago

Read 2 more answers

Olympic athletes have lower resting heart rates because

Burka [1]

<span>they exercise regularly, which improves the efficiency of their heart.

Hope this helps!

-Payshence xoxo</span>

3 0

3 years ago

Germanium is a group 4A semiconductor. The addition of a dopant atom (group 3A element) that has fewer valence electrons than th

klasskru [66]

Answer:

Doping with galium or indium will yield a p-type semiconductor while doping with arsenic, antimony or phosphorus will yield an n-type semiconductor.

Explanation:

Doping refers to improving the conductivity of a semiconductor by addition of impurities. A trivalent impurity leads to p-type semiconductor while a pentavalent impurity leads to an n-type semiconductor.

7 0

2 years ago

) PABA refers to para-aminobenzoic acid which is used in some sunscreen formulations. If a 0.055 M solution of PABA has a pH of

AleksandrR [38]

Answer: $K_a$ of PABA is 0.000022

Explanation:

$NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.055 M and $\alpha$ = ?

$K_a=?$

$pH=-log[H^+]$

$2.96=-log[H^+]$

$[H^+]=1.09\times 10^{-3}$

$[H^+]=c\times \alpha$

$1.09\times 10^{-3}=0.055\times \alpha$

$\alpha=0.02$

Putting in the values we get:

$K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}$

$K_a=0.000022$

Thus $K_a$ of PABA is 0.000022

5 0

2 years ago