Solutions of sodium sulfate and silver nitrate are mixed together.
On a piece of scratch paper write a fully balanced double-replacement equation for this reaction.
Using the chemical equation you wrote; which answer correctly identifies the precipitate (if there is one)
and the net ionic equation for the reaction that formed the precipitate?
There is no precipitate for the reaction; all products are soluble.
Precipitate: silver sulfate
Net lonic: 2Ag+ + SO42- --> Ag2SO4
Precipitate: sodium nitrate
Net lonic: Na2+ + 2NO3 --> Na(NO3)2
Precipitate: silver sulfate
Net lonic: Ag+ + SO42- --> AgSO4
Precipitate: sodium nitrate
Net lonic: Na+ + NO3 --> NaNO3

Precipitate is silver sulfate and the net ionic equation is

2Ag⁺(aq) + SO₄²⁻(aq) → Ag₂SO₄(s)

Explanation:

The given reaction is the reaction of sodium sulfate with silver nitrate and as the reaction goes we will get the precipitate of Silver sulfate and the balanced reaction is given as,

Na₂SO₄ + 2 AgNO₃→ Ag₂SO₄↓ +  2 NaNO₃

Here silver sulfate is obtained as a precipitate.

Net ionic equation is the equation in which the ions involved in the reaction only taken into account that is ions forming precipitate, whereas the other ions are considered as spectator ions, ions do not involved in the reaction.

Here the net ionic equation is:

2Ag⁺(aq) + SO₄²⁻(aq) → Ag₂SO₄(s)

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'