True or false nothing is smaller than atom

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

2 years ago

ammonia gas is used as refrigerant 0.474 atm. Pressure is required to change 2000 cm3 sample of ammonia initially at 1.0 atm to

Andreas93 [3]

Answer: The pressure required is 0.474 atm

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}V}$ (At constant temperature and number of moles)

The equation is,

${P_1V_1}={P_2V_2}$

where,

$P_1$ = initial pressure of gas = 1.0 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = $2000cm^3$

$V_2$ = final volume of gas = $4.22dm^3=4220cm^3$ ( $1dm^3=1000cm^3)$

Now put all the given values in the above equation, we get:

${1.0\times 2000}={P_2\times 4200}$

$P_2=0.474atm$

The pressure required is 0.474 atm

5 0

2 years ago

Which of the following do you pracrice at home

S_A_V [24]

You didn’t post the photo or full question

8 0

2 years ago

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of water i

Kaylis [27]

Answer:

87.9%

Explanation:

Balanced Chemical Equation:

HCl + NaOH = NaCl + H2O

We are Given:

Mass of H2O = 9.17 g

Mass of HCl = 21.1 g

Mass of NaOH = 43.6 g

First, calculate the moles of both HCl and NaOH:

Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles

Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles

Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:

Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles

Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles

From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:

Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g

% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%

3 0

3 years ago

3. A 4.00 gram sample of solid gold was heated from 274K to 314K. How much energy was involved?

MrMuchimi

Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).

3 0

2 years ago