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qaws [65]
3 years ago
10

How does each cell accomplish a wide variety of activities?

Chemistry
2 answers:
VLD [36.1K]3 years ago
8 0
The organelles have specialized functions, (D)
FromTheMoon [43]3 years ago
8 0
D. because the organelles could help the cell breath or survive in certain areas 
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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
Consider the following equilibrium systems: A⇌2B ΔH o =20.0 kJ/mol Reaction 1 A+B⇌C ΔH o =−5.4 kJ/mol Reaction 2 A⇌B ΔH o =0.0 k
galina1969 [7]

Answer:

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.

8 0
3 years ago
Whats the answer to 7/2 × 1/5?​
svp [43]
The answer to the problem is 7/10
3 0
3 years ago
Read 2 more answers
Nuclear reactions involves​
Finger [1]
Nuclear reaction involves two reacting particles a heavy target nucleus and a light bombarding particle and produces two new particles a heavier product nucleus and a lighter ejected particle.
6 0
3 years ago
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.
timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU +  w

q = 475 J  +  93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0
3 years ago
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