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Arisa [49]
3 years ago
11

How many electrons should Nitrogen (N) have around its Lewis dot model?

Chemistry
1 answer:
Delvig [45]3 years ago
3 0

Answer:

5

Explanation:

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How many mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound?
saul85 [17]

2.87mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a substance can be calculated by dividing the number of moles of the substance by its molarity. That is;

volume = no. of moles ÷ molarity

According to this question, 2.5M K2SO4 are required to obtain 1.25 grams of the compound. The volume can be calculated as follows:

Molar mass of K2SO4 = 174.26 g/mol

moles of K2SO4 = 1.25g ÷ 174.26g/mol

moles of K2SO4 = 0.00717moles

Volume = 0.00717moles ÷ 2.5

Volume = 0.00287 L = 2.87mL.

Therefore, 2.87mL of 2.5 M K2SO4 are required to obtain 1.25 grams of the compound.

Learn more about volume at: brainly.com/question/1578538

6 0
2 years ago
What is the fusing inside the sun to make helium
lawyer [7]
By nuclear fusion of hydrogen nuclei into Helium
6 0
3 years ago
Read 2 more answers
We want to find how much charge is on the electrons in a nickel coin. follow this method. a nickel coin has a mass of about 5 g.
Lady bird [3.3K]
The mass of a nickel coin is 5 g.
1 mol of Ni weighs 58 g. 1 mol contains 6.022 x 10²³ atoms of Ni.
therefore in 58 g  there are 6.022 x 10²³ atoms of Ni
then in 5 g the number of Ni atoms are - 6.022 x 10²³ /58 x 5 = 5.2 x 10²² Ni atoms
Therefore number of Ni atoms are 5.2 x 10²² atoms in a nickel coin

3 0
3 years ago
1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

6 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
3 years ago
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