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sergiy2304 [10]
3 years ago
7

Locate on the periodic table each group of ions with the same charge. What relationship exists between ions of the same charge a

nd the location of the corresponding element on the periodic table?​
Chemistry
1 answer:
DiKsa [7]3 years ago
4 0

Answer:

The periodic table is a tabular display of the chemical elements organized on the basis of their atomic numbers, electron configurations, and chemical properties. ... Elements with the same number of valence electrons are kept together in groups, such as the halogens and the noble gases.

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D

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From the following balanced equation,
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9.72 grams.

Explanation:

From the equation, 4 moles of NH₃ produce 6 moles of  water.

Therefore the reaction to product ratio of  NH₃ to H₂O is 4:6

and 2:3 into its simplest form.

The number of moles of NH₃ in 6.12 g is:

Number of moles=mass/ RMM

=6.12 g/17 G/mol

=0.36 moles.

Therefore the number of moles of H₂O produced is calculated as follows.

(0.36 Moles×3)2 = 0.54 moles

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6 0
3 years ago
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
3 years ago
GIVING BRAINLIEST AND THE REST OF MY POINTS!!!!! :D
Alik [6]
It’s d I did this before
6 0
3 years ago
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