Mr: 207.2
m=n×Mr= 6.53×207.2= 1353.02g
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
Explanation:
A. Linear Relationships
x 2 4 6 8 10 12
y 4.4 8.8 13.2 17.6 22 26.4
(you can plot this by using the (x,y) coordinates for each pair above.)
B. y=2.2x
mass is 2.2 times larger than the volume.
C. The mass is 2.2 times the volume.
<span>The number of neutrons bromine will have are equal to
= protons + neutrons
so,
80-35=45</span>
<u>Answer:</u> The volume when the pressure and temperature has changed is 
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
Let us assume:
![P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K](https://tex.z-dn.net/?f=P_1%3D1.20atm%5C%5CV_1%3D795mL%5C%5CT_1%3D116%5EoC%3D%5B116%2B273%5DK%3D389K%5C%5CP_2%3D0.55atm%5C%5CV_2%3D%3FmL%5C%5CT_2%3D75%5EoC%3D%5B75%2B273%5DK%3D348K)
Putting values in above equation, we get:
Hence, the volume when the pressure and temperature has changed is
