Balance the equation (nh4)3 po4 +na0h arrow na3p04 +3nh3 +3h20.

Explanation:

(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O

Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH

(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O  Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.

You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.

There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.

Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.

Species      Na        H        O         N       PO4

Left             3          15        3         3          1

Right           3         15         3         3          1

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

Explanation:

Your unbalanced equation is

(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

A method that usually works for balancing by inspection is

1. Pick the most complicated-looking formula [(NH₄)₃PO₄].

Put a 1 in front of it.

1(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

2. Balance N.

We have fixed 3N on the left. We need 3N on the right.

Put a 3 in front of NH₃.

1(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + 3NH₃ + H₂O

3. Balance P.

We have fixed 1P on the left. We need 1P on the right.

Put a 1 in front of Na₃PO₄.

1(NH₄)₃PO₄ + NaOH ⟶ 1Na₃PO₄ + 3NH₃ + H₂O

4. Balance Na

We have fixed 3Na on the right. we need 3Na on the left.

Put a 3 in front of NaOH.

1(NH₄)₃PO₄ + 3NaOH ⟶ 1Na₃PO₄ + 3NH₃ + H₂O

5. Balance O.

We have fixed 7O on the left and 4O on the right. We need three more O atoms on the right.

Put a 3 in front of H₂O.

1(NH₄)₃PO₄ + 3NaOH ⟶ 1Na₃PO₄ + 3NH₃ + 3H₂O

All species have a coefficient. The equation should now be balanced.

6. Check that all atoms are balanced

The balanced equation is

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%