All categories

3 years ago

Problem 2 - The mass of Earth is 6.0x1024 kg. The mass of a hydrogen atom is

Chemistry

1 answer:

n200080 [17]3 years ago

4 0

Answer: has diffused more quickly, so it must be a

lighter gas. Its particles have greater velocity

than the particles of Y at the same temperature.

(Note though that they will both have the same

value for average kinetic energy.)

7 From the kinetic molecular theory we would

expect a solid to be more dense than its liquid,

and therefore that ice would sink in water.

8 Bubbles will be present through the volume

of the liquid. A brown gas is visible above the

brown liquid. As the two states are at the same

temperature, the particles have the same average

kinetic energy and are moving at the same

speed. The inter-particle distances in the gas are

significantly larger than those in the liquid.

9 At certain conditions of low temperature and low

humidity, snow changes directly to water vapour

by sublimation, without going through the liquid

phase.

10 Steam will condense on the skin, releasing energy

as it forms liquid at the same temperature (E–D

on Figure 1.4). This is additional to the energy

released when both the boiling water and the

condensed steam cool on the surface of the

Explanation:

You might be interested in

If an atom has 15 protons, 11 neutrons, and 16 electrons, what is the atom's electrical charge? A. -5 B. -1 C. +1 D. +5

Helen [10]

Answer:

it's out of the syllabus I can't help

Explanation:

Blanka ajabdjdhsjbsskhsjshsjsvssgshshs

4 0

2 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0

3 years ago

Plants undergo photosynthesis to produce glucose according to the reaction below. What mass of water is required to produce 5.0g

solniwko [45]

Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

6 mol 1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆ = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

6CO₂ + 6H₂O ---------> C₆H₁₂O₆ + 6O₂

6 mol (18 g/mol) 1 mol (180 g/mol)

108 g 180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

mass of water (H₂O) = 108 g x 5 g / 180 g

mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.

7 0

3 years ago

If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume

fomenos

Answer:

Explanation:

A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.

The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.

The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.

5 0

3 years ago