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3 years ago

The number of bonded pairs of _______ in a covalent bond equals the bond order.

1 answer:

5 0

The number of bonded pairs of electrons in a covalent bond equals the bond order.

For example:
In molecule of hydrogen (H₂), bond order is 1 (H-H), because there is one pair of electrons between hydrogen atoms.
In molecule of oxygen (O₂), bond order is 2 (O=O), because there are two pair of electrons (4 electrons) between oxygen atoms.
In molecule of nitrogen (N₂), bond order is 3 , because there are three pair of electrons (6 electrons) between nitrogen atoms.

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An insoluble base reacts with an acid to form a salt

Alex17521 [72]

Answer: you can use a strainer

Explanation:

4 0

3 years ago

This is my chemistry worksheet. It's a new topic my teacher will barely review with us. Help please???

TEA [102]

So to balance an equation, you need to get the same amount of each type of element on either side of the --> . So you pretty much are given the subscripts in the equations and you need to add coefficients (just normal numbers) in front of any formula that needs it, keeping anything balance.
$KCl_{3} + O_{2} -\ \textgreater \ KCl_{3}$
turns into
$2KCl_{3}+ 3O_{2} -> 2KCl_{3}$

These coefficient numbers are the molar ratios, so 2 moles of KCl3 for every 3 moles of O2 so 1. 3:2

Then you can use these ratios of find out how many moles of one thing are needed if you are given the amount of another.
$\frac{moles of element 1}{cofficient 1} = \frac{moles of element 2}{cofficient 2}$
and use cross multiplication to solve for whatever you don't know

6 0

3 years ago

19. What is sthe mass of 6.02 x 1024 atoms

alekssr [168]

The mass of magnesium in $6.02 \times 10^{24}$ atoms is 240 g.

Answer: Option A

Explanation:

First, we have to convert the atoms to moles of magnesium.

We know that $6.02 \times 10^{24}$ atoms are present in 1 mole of magnesium. So,

$\text { 1 atom }=\frac{1}{6.02 \times 10^{24}} \text { moles }$

$6.02 \times 10^{24} \text { atoms }=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}} \text { moles }=10 \text { moles }$

Thus,

$Mass of M g= molar mass of M g \times No. of moles of M g$

$\text { Mass of } M g=24 \times 10=240 g$

Thus, 240 g of Magnesium is present in $6.02 \times 10^{24}$ atoms.

4 0

3 years ago

Hydrochloric acid reacts with calcium to form hydrogen and calcium chloride. if 100 grams of hydrochloric acid reacts with 100 g

natita [175]

The balanced chemical reaction is:

2HCl + Ca = CaCl2 + H2

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

100 g HCl ( 1 mol HCl / 36.46 g HCl ) = 2.74 mol HCl

100 g Ca ( 1 mol Ca / 40.08 g ) = 2.08 mol Ca

From the reaction, the mole ratio of the reactants is 2:1 where every 2 moles of hydrochloric acid, 1 mole of calcium is required. Therefore, the limiting reactant for this case is calcium.

4 0

3 years ago

Read 2 more answers

How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

zaharov [31]

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!

7 0

3 years ago