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tiny-mole [99]
3 years ago
11

Provide the name of the compound shown below. Spelling and format counts!CH3NO2NO2​

Chemistry
1 answer:
lesantik [10]3 years ago
4 0

Answer:

ch3:-methanide

No2:-nitrogen dioxide

thank you and mark as brainliest if u find it helpful

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Pls help, How is stoichiometry used to calculate amounts of substances in a chemical reaction?
Elden [556K]

Answer:

Thus, to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.

Explanation: Stoichiometry is the field of chemistry that is concerned with the relative quantities of reactants and products in chemical reactions. For any balanced chemical reaction, whole numbers (coefficients) are used to show the quantities (generally in moles ) of both the reactants and products.

7 0
3 years ago
Which sample has particles with the lowest average kinetic energy?
Anastasy [175]
The question asks average kinetic energy. So it is only related with the temperature. The higher temperature is, the higher kinetic energy is. So the answer is (4).
3 0
3 years ago
Read 2 more answers
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.56 mol of m
pishuonlain [190]
<h3>Answer:</h3>

43.27 g Mg

<h3>Explanation:</h3>

The balanced equation for the reaction between magnesium metal and hydrochloric acid is;

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

From the equation;

1 mole of magnesium reacts with 2 moles of HCl

We are given;

3.56 moles of Mg and 3.56 moles of HCl

Using the mole ratio;

3.56 moles of Mg would react with 7.12 moles of HCl, and

3.56 moles of HCl would react with 1.78 moles of Mg

Therefore;

The amount of magnesium was in excess;

Moles of Mg left = 3.56 moles - 1.78 moles

                         = 1.78 moles

But; 1 mole of Mg = 24.305 g/mol

Therefore;

Mass of magnesium left = 1.78 moles × 24.305 g/mol

                                        = 43.2629 g

                                        = 43.27 g

Thus, the mass of magnesium that remained after the reaction is 43.27 g

4 0
3 years ago
Given 5.0 grams of lead (II) nitrate and 3.0
natulia [17]

Answer:

                     1.70 g of NaNO₃

Explanation:

                     The balance chemical equation for given double displacement reaction is,

                               Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: <u>Calculate moles of each reactant:</u>

                                   Moles =  Mass / M.Mass

For Pb(NO₃)₂:

                                   Moles =  5.0 g / 331.21 g/mol

                                   Moles =  0.0150 moles

For NaI:

                                   Moles =  3.0 g / 149.89 g/mol

                                   Moles =  0.020 moles

Step 2: <u>Calculate Limiting reagent as;</u>

According to equation,

                     1 mole of Pb(NO₃)₂ reacts with  =  2 moles of NaI

So,

              0.0150 moles of Pb(NO₃)₂ will react with  =  X moles of NaI

Solving for X,

                     X =  2 mol × 0.0150 mol / 1 mol

                     X =  0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: <u>Find out moles of NaNO₃ formed:</u>

According to equation,

                     2 mole of NaI produced =  2 moles of NaNO₃

So,

              0.020 moles of NaI will produce   =  X moles of NaNO₃

Solving for X,

                     X =  2 mol × 0.020 mol / 2 mol

                     X =  0.020 moles of NaNO₃

Step 4: <u>Calculate Mass of NaNO₃ produced;</u>

As,                Moles  =  Mass / M.Mass

Or,

                    Mass  =  Moles × M.Mass

Putting values,

                    Mass  =  0.020 mol × 84.99 g/mol

                    Mass  =  1.70 g of NaNO₃

7 0
3 years ago
How many grams of water can be produced when 8.73 moles of benzene (C6H6) react with excess oxygen?. . Unbalanced equation: C6H6
Alex_Xolod [135]
<span>2 C6H6 + 15 O2 = 12 CO2 + 6 H2O
</span>
2 moles C6H6 -------------------- 6 moles H2O
8.73 moles C6H6 --------------- ( moles H2O)

( moles H2O) = 8.73 x 6 / 2

( moles H2O) = 52.38 / 2

= 26.19 moles of H2O

Molar mass H2O = 18.00 g/mol

1 mole -------------- 18.00 g
26.19 moles ----- ( mass of H2O)

mass of H2O = 26. 19 x 18.00 => 471.42 g of water

hope this helps!


7 0
3 years ago
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