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3 years ago

A sample of argon has a volume of 5.0 dm^3 and the pressure is 0.92 atm.

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

5 0

Answer:

232.9K

Explanation:

V₁ = 5dm³

P₁ = 0.92atm

T₂ = 30°C

V₂ = 5.7L

P₂ = 800mmHg

Unknown:

T₁ = ?

Solution:

To solve this problem, we are going to apply the combined gas law which is fusion of the avogadro's law, boyle's law and charles's law.

Mathematically;

$\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }$

p, v and T are pressure, volume and temperature respetively

1 and 2 depicts the initial and final states.

We need to convert ;

temperature to K and pressure into atm

T₂ = 30°C = 273 + 30 = 303K

P₂ = 800mmHg;

760mmHg = 1 atm

800mmHg = $\frac{800}{760}$ = 1.05atm

Input the variables and solve;

$\frac{0.92 x 5}{T} = \frac{1.05 x 5.7}{303}$

T = 232.9K

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miss Akunina [59]

Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

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pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

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5 0

3 years ago

The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat

Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

Rate constant for first order reaction is as follows.

$k_{d} = \frac{0.6932}{480}$

= $1.44 \times 10^{-3}s^{-1}
$

Initial fat concentration $S_{o}$ = 45 $mol/m^{3}$

= 45 mmol/L

Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = $S_{o} (1 - X)$

= 45 (1 - 0.80)

= 9 $mol/m^{3}$

or, = 9 mmol/L

It is given that $K_{m}$ = 5mmol/L

Therefore, time taken will be calculated as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]$

= $1642.83 s \times \frac{1 min}{60 sec}$

= 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0

3 years ago

Plz someone help, really struggling

frez [133]

Answer:

22.9 Liters CO(g) needed

Explanation:

2CO(g) + O₂(g) => 2CO₂(g)

? Liters 32.65g

= 32.65g/32g/mol

= 1.02 moles O₂

Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)

∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)

Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.

∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed

___________________

*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).

6 0

3 years ago

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Reil [10]

Answer:

A is the correct option

Explanation:

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4 0

3 years ago

Which cup has more thermal energy?

dybincka [34]

Question: Which cup has more thermal energy?

A. 2 liter cup of milk

B. 1 liter cup of milk

Answer: A. 2 liter cup of milk

Explanation:

We have milk in both of the containers. The milk in both containers is the same temperature, so we will have to measure the amount of molecules. The amount of molecules makes it warmer, and if it is warmer, that is more energy. 2 liters have more thermal energy than 1 liter, because it has more molecules.

Our answer is 2 liter cup of milk.

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3 years ago