All categories

3 years ago

How do you balance the equation in number one?

Chemistry

1 answer:

AleksAgata [21]3 years ago

3 0

____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O

To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.

To start off, you wanna know the number of atoms in each element on both sides, so take it apart:

[reactants] [product]
Na- 1 Na- 2

N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)

O- 4 O- 7 (same reason as above)

Pb- 1 Pb- 1

Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:

2NaNO3

Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:

Na- 2 Na- 2

N- 2 N- 2 (it balanced out)

O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)

Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).

Now to put it back together, it will look like this:

2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O

You might be interested in

What is the nuclear binding energy of an atom that has a mass defect of 5.0446

notsponge [240]

Answer:

Option C: 4.54 x $10^{11}$ KJ/mol of nuclei

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$ .

Explanation:

If mass defect is known, then nuclear binding energy can easily be calculated, here's how:

First step is to convert that mass defect into kg.

Mass defect = 5.0446 amu

Mass defect = 5.0466 x 1.6606 x $10^{-27}$

Because 1 amu = 1.6606 x $10^{-27}$ Kg.

Mass defect = 8.383 x $10^{-27}$ kg.

Now, we need to find out it's energy equivalent by using following equation:

Using the equation E = mc²:

where c= 3.00 x $10^{8}$ m/s²

E = (8.383 x $10^{-27}$ ) x (3.00 x $10^{8}$ )²

E = 7.54 x $10^{-10}$ J this energy is in Joules but nuclear binding energy is usually expressed in KJ/mol of nuclei. Let's convert it:

(7.54 x $10^{-10}$ Joule/nucleus)x(1 kJ/1000 Joule)x(6.022 x $10^{23}$ nuclei/mol) =

4.54 x $10^{11}$ kJ/mol of nuclei .

E = 4.54 x $10^{11}$ kJ/mol of nuclei . So, this is the nuclear binding energy of that atom, which is option C.

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$

4 0

3 years ago

Read 2 more answers

Why can't air exist in space, but gas can????

MakcuM [25]

Answer:

Earth's gravity is strong enough to hold onto its atmosphere and keep it from drifting into space.

8 0

2 years ago

1.* Write the name and symbol of the ion formed when

wlad13 [49]

1a) anion and S-2
1b) cation and Al+3
1c) anion and N-3
1d) cation and Ca+2

6 0

3 years ago

Read 2 more answers

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

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

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

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

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

4 0

3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

2 years ago