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3 years ago

19. When 0.5g of calcium trioxocarbonate (IV) was added to excess dilute

Chemistry

1 answer:

Flauer [41]3 years ago

7 0

Answer:

0.1gmin-1

Explanation:

mass = 0.5g

time = 5minutes

rate= mass/time = 0.1gmin-1

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What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 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

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution

natulia [17]

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume = 100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

7 0

3 years ago

Which of the following are NOT properties of an base

Sav [38]

Answer:

Reacts with salt to from an acid

6 0

3 years ago

How many grams of Aluminum are needed to<br> produce 45.3 grams of Iron metal?

Dimas [21]

It might be 90.6g...

3 0

3 years ago