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Alborosie
2 years ago
3

Calculate the mass percent of calcium in calcium carbonate. Report your answer to at least 3 significant figures. Hint: This cal

culation is theoretical, and is independent of the data provided. Another way to phrase the question would be: What percentage of the molar mass of CaCO3 is due to the Ca?
Chemistry
1 answer:
lesya692 [45]2 years ago
4 0

Answer:

40%

Explanation:

Molar  mass  of Ca = 40 gram/mol

Molar mass of C      =  12 gram/mol

Mass mass of O =  16 gram/mol

Molar mass of CaCO3 = [40 + 12 + (16 x3)]

                                     = 100 gram/mol

Mass % of Calcium = Ca/ CaCO3

                                =   40/100

                                = 40%

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A plot of velocity versus substrate concentration for a simple enzyme-catalyzed reaction produces a _____. This indicates that a
kirza4 [7]

Answer:

B) hyperbolic curve; saturated with substrate

Explanation:

Enzymatic kinetics studies the speed of enzyme catalyzed reactions. These studies provide direct information about the mechanism of the catalytic reaction and the specificity of the enzyme. The speed of a reaction catalyzed by an enzyme can be measured with relative ease, since in many cases it is not necessary to purify or isolate the enzyme. The measurement is always carried out under the optimal conditions of pH, temperature, presence of cofactors, etc., and saturating substrate concentrations are used. Under these conditions, the reaction rate observed is the maximum speed (Vmax). The speed can be determined either by measuring the appearance of the products or the disappearance of the reagents.

Following the rate of appearance of product (or disappearance of the substrate) as a function of time, the so-called reaction progress curve is obtained, or simply, the reaction kinetics. This curve is represented by a hyperbolic curve

5 0
2 years ago
Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125
expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0
2 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
What is the mass of 22.4 L of H2 at STP?
Vanyuwa [196]

A. 1.01 is the right answer

Since

The formula is Pv= nRT

P=1 atm

V= 22.4 L

N= x

r= 0.0821

t = 273 k (bc it’s standard temperature)

So (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0
2 years ago
Read 2 more answers
Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?
Fantom [35]

Answer:

Option D is correct.

H₂O + CO₂      →    H₂CO₃

Explanation:

First of all we will get to know what law of conservation of mass states.

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Example:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Now we will apply this law to given chemical equations:

A) H₂ + O₂   →    H₂O

There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.

B) Mg + HCl   →   H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃      →     KCl + O₂

There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D)  H₂O + CO₂      →    H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.

6 0
2 years ago
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