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Reil [10]
3 years ago
13

Hellllllppppppp!!!!!!!! 5 more minnnnnnn C. To produce D. To have a purpose

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
6 0
I believe it is C) to produce
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I need help on this, It’s due today. Please, anybody
Sholpan [36]

Answer:

1 = 252g, 2 = 2mL, 3 = 1.5mL, 4 = 3g, 5 = 225g, 6 = 0.92g/mL, 7 = 0.75g/mL, 8 = 0.71g/mL, 9 = 1.9mL, 10= 1.11mL, 11 = 76.9g

Explanation:

This problem is testing how well you can move around the equation D = m/v where D = Density (g/mL), m= mass of sample (g), v = volume of sample (mL).

4 0
3 years ago
Which best explains why a crystal is incompressible?.
Katen [24]

Answer:

There is little if any, space left between its adjacent molecules.

Good luck!

6 0
2 years ago
Why are enzymes considered catalyst?
ss7ja [257]
A catalyst<span> is a chemical that increases the rate of a chemical reaction without itself being changed by the reaction. The fact that they aren't changed by participating in a reaction distinguishes </span>catalysts<span> from substrates, which are the reactants on which </span>catalysts<span> work. </span>Enzymes<span> catalyze biochemical reactions.</span>
8 0
3 years ago
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

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

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

3 0
3 years ago
What evidence should Mitchell collect to BEST support the second law of thermodynamics?
Dominik [7]

Answer:

The Second Law of Thermodynamics poses an insurmountable problem for ... More time will make things worse for the Darwinist, not better. ... error fail-safe and proof-reading devices utilized for quality control, assembly ... No one really supports the idea that the sun + non life = life

Explanation:

6 0
2 years ago
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