Answer:
39.4 kJ
Explanation:
Given mass = 25.0 g
Specific heat of ammonia (g) = 4.7 J/g.K
Specific heat of ammonia (l) = 2.2 J /g.K
ΔHvap = 23.5 x 10³ J /mol
Heating liquid ammonia from -65 °C to -33.4°C
Q₁ = m s ΔT = = 25.0 g * 4.7 J/g.K * [-33.4 - (-65)] °C = 3713 J
Evaporating 25 g of ammonia at -33.4°C
Molar mass of ammonia = 17 g/mol
Thus, moles = 25 / 17 moles
Q₂ = (25 / 17) mol * ΔHvap = 1.470 mol * 23.5 x 10³ J /mol = 34558.8 J
Heating NH₃(g) from -33.4°C to -12.0°C
Q₃ = m s ΔT = = 25.0 g * 2.2 J/g.K * [-12.0 - (-33.4)] °C = 1177 J
<u>Total heat energy = Q₁ + Q₂ +Q₃ = 39448.8 J = 39.4 kJ (As 1 J = 10⁻³ kJ)</u>
Any lump of osmium is. Iridium and gold are also very close.
sign out and log in again...if does not work then make a new account
Answer:
w=fx
w=(15)(5)
w=75
the angle between the deplacement and N is 0
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b) 
And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:

And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10