What is the percent composition of (NH4)2S? if you can, include the problem shown pls
extra.
Mg(NO3)2
please don't just take the points please :(, do the problem
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(NH₄)₂S

%N=  41.18%

%H =11.76%

%S=47.06%

Mg(NO₃)₂

%Mg=32.43%

%N=18.92%

%O=(64.86%

Given

(NH₄)₂S

Mg(NO₃)₂

Required

The percent composition

Solution

(NH₄)₂S MW=68 g/mol

%N=2. Ar N  / MW (NH₄)₂S x 100%

%N= (2.14/68) x 100% = 41.18%

%H =(8.1/68) x 100%=11.76%

%S=(32/68)x 100%=47.06%

Mg(NO₃)₂ MW=148 g/mol

%Mg=(2.24/148)x100%=32.43%

%N=(2.14/148)x100%=18.92%

%O=(6.16/148)x100%=64.86%

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