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3 years ago

5

The parts of a mixture O A) form chemical bonds with each other. O B) combine to form new compounds. O c) cannot be physically s

eparated O D) show their individual properties

1 answer:

katrin2010 [14]3 years ago

7 0

Answer:B

Explanation:

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Determine the volume occupied by 0.352 mole of a gas at 25⁰C if the pressure is 81.8 kPa.

lions [1.4K]

Data:
p (pressure) = 81.8 kPa = 81.8*10³ Pa ≈ 8.07 atm
v (volume) = ? (in L)
n (number of mols) = 0.352 mol
R (Gas constant) = 0.082 (atm*L/mol*K)
T (temperature) = 25ºC converting to Kelvin, we have:
TK = TC + 273 → TK = 25 + 273 → TK = 298

Formula:
$p*V =n*R*T$

Solving:
$p*V =n*R*T$
$8.07*V = 0.352*0.082*298$
$8.07V \approx 8.60$
$V \approx \frac{8.60}{8.07}$
$\boxed{\boxed{V \approx 1.06\:L}}$

7 0

3 years ago

To make sure that the white powder was all sodium chloride and not mixed with sodium hydrogen carbonate, would you need to add a

TiliK225 [7]

Answer: i hope you thought this was an actual answer

Explanation:

6 0

2 years ago

Examine the given reaction.

melamori03 [73]

Answer:

The correct option is;

D)

Explanation:

The given reaction is presented as follows;

NH₄Cl (s) → NH₃ (g) + HCl (g) ΔH° = 176 kJ/mol, ΔS° = 0.285 kJ/(mol·K)

We note that the Gibbs free energy, ΔG° is represented by the following equation;

ΔG° = ΔH° - T·ΔS°

Where:

T = Temperature (Kelvin)

The reaction will be spontaneous for exergonic reactions, ΔG° < 0 and it will not be spontaneous for endergonic reaction, ΔG° > 0

At room temperature, T = 25 + 273.15 = 298.15 K

Which gives;

ΔG° = 176 - 298.15 × 0.285 = 91.03 kJ/mol which is > 0 Not spontaneous reaction

At 800°C, we have;

T = 273.15 + 800°C + 1073.15 K

ΔG° = 176 - 1073.15 * 0.285 = -129.85 kJ/mol which is < 0 the reaction will be spontaneous

The correct option is therefore, that at room temperature, the reaction is not spontaneous. However, at high temperatures. like 800 °C, the free energy value turns negative and this reaction becomes spontaneous.

4 0

3 years ago

When 131.0 mL of water at 26.0°C is mixed with 81.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat i

JulsSmile [24]

Answer:

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m}{131.0 mL}$

m = 131.0 g

Initial temperature of the water = $T_i$ = 26.0°C

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i$

Heat absorbed 131.0 g of water = Q

$Q=m\times c\times \Delta T$

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m'}{81.0 mL}$

m' = 81.0 g

Initial temperature of the water = $T_i'$ = 85.0°C

Final temperature of the water = $T_f'$

Change in temperature , $\Delta T'=T_f'-T_i'$

Heat lost by 81.0 g of water = Q'

$Q'=m'\times c\times \Delta T'$

After mixing both liquids the final temperature will become equal fro both liquids.

$T_f=T_f'$

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

$m\times c\times \Delta T=m'\times c\times \Delta T'$

$131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))$

$T_f=63.52^oC$

63.52°C is the final temperature

4 0

3 years ago

Zach needs to make a 2.75 m solution of NaOH(molar mass=40.00g/mol) using 350. g of water. What mass of NaOH should he dissolve

goldfiish [28.3K]

Molality = number of mol solute/ 1000 g solvent = (2.75 mol NaOH)/1000g water

$\frac{2.75( mol NaOH)}{1000(g/ water)} = \frac{x (mol NaOH)}{350(g/water)} 


$ ,
$x= \frac{2.75*350}{1000} = 0.963 (mol NaOH)

$ $0.963 (mol NaOH)* \frac{40.00 (g NaOH)}{1 ( mol NaOH)} =38.5 (g NaOH)$

3 0

3 years ago