Answer:
A. 1350
You multiply 18.21HNO3* 1mol MgN2O6 * 148.30MgN2O6
Then divide it by the 2mol HNO3 to get 1350
I think it’s chemical reactivity
Answer:
The length of foil will be 8107.81 cm or 81.7081 m.
Explanation:
Given data:
Width of roll of foil = 302 mm
Height or thickness = 0.018 mm
Density of foil = 2.7 g/cm³
Mass of foil = 1.19 Kg
Length of foil = ?
Solution:
d = m/ v
v = length (l) × width (w) × height (h)
First of we will convert the Kg into gram and mm into cm.
one Kg = 1000 g
1.19 × 1000 = 1190 g
one cm = 10 mm
302 / 10 = 30.2 cm
0.018 / 10 = 0.0018 cm
Now we will put the values in formula:
d = m/ l× h× w
l = m / d × h× w
l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm
l = 1190 g/ 0.146772 g/cm
l = 8107.81 cm or 81.7081 m
Answer:
the 3rd one (0.01 cm the one selected already)
Explanation:
copper wire isn't excessively big, and it wraps around the pencil because its malleable. I think that the most accurate would be 0.01 cm
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
