Answer:
NH4Br + AgNO3 —> AgBr + NH4NO3
Explanation:
When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:
NH4Br(aq) + AgNO3(aq) —>
In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:
NH4Br(aq) —> NH4+(aq) + Br-(aq)
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
The double displacement reaction will occur as follow:
NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)
NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)
Answer:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
Explanation:
This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:
HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)
However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:
2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)
Answer:
<u>136.67 g of Na3PO4 i</u>s required to create 100 gram of NaOH.
Explanation:
The balanced equation:

1 mole Na3PO4 = 164 g/mole (Molar mass)
1 mole NaOH = 40 g/mole (Molar mass)
Now,
1 mole of Na3PO4 produce = 3 mole of NaOH
164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH
or
120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4
1 g/mol of NaOH is produced from =

100 grams of NaOH is produced from =
gram of Na3PO4
calculate,
= 136.67 g
Answer:
C: liquid
Explanation:
Liquid has the set volume as it cannot be compressed or 'expanded' like a gas. However, liquids will take the shape of the container you put them in so their shape can change.