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2 years ago

Please help me balance:_NaNO3 + _PbO_ ➡️ Pb(NO3)2 + _Na2O

Chemistry

1 answer:

KengaRu [80]2 years ago

7 0

Answer:

The balanced reaction is given by,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

Explanation:

The reaction is as given.

Lets count the number of each elements in the reaction.

In reactant side, number of sodium atoms are 1 , lead are 1, nitrogen are 1 and oxygen are 4.

in product side, number of sodium atoms are 2 , lead are 1 , nitrogen are 2 and oxygen are 7.

So we need to balance sodium and oxygen atoms in the reaction.

There is deficient of sodium and oxygen atoms on reactant side.

Thus, multiply (NaNO3) by 2.

Thus, sodium atoms become 2 , nitrogen 2 and oxygen 6. Total 7 oxygen atoms.

Thus, the balanced reaction is,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

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Read 2 more answers

How many grams of sodium oxide can be synthesized from 17.4 g of sodium? assume that more than enough oxygen is present. the bal

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Equation is as follow,

 4 Na (s) + O₂ (g) → 2Na₂O (s)

According to equation,

91.92 g (4 moles) of Na produces = 123.92 g (2 moles) of Na₂O
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3 years ago

Mature elephants consume an average of 600 pounds of food per day. If an elephant

MAXImum [283]

Answer:

$\large \boxed{\text{4 600 000 kg or 4600 Mg}}$

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

$\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}$

3. Convert days to pounds of feed

$\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}$

4. Convert pounds to kilograms and megagrams

$\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}$

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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3 years ago

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

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2 years ago