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pshichka [43]
3 years ago
5

Hi! I need some help with this pleaseeee!!

Chemistry
1 answer:
german3 years ago
6 0
A) eventually they will be in thermal equilibrium
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One of the following is not a factor which affect rates of reaction:
kotegsom [21]

Answer:

d. Pressure

Explanation:

Pressure is not a factor which affect rates of reaction.

However, surface area, temperature, and concentration are factors that affect rates of reaction.

8 0
3 years ago
1. What is the density, in g/cm², of a substance weighing 22.2 cg with dimensions of 0.333
RUDIKE [14]
  • Mass=22.2cg=2.22g

Volume:-

\\ \tt\longmapsto 0.333(0.444)(0.555)=0.08cm^3

So

\\ \tt\longmapsto Density=\dfrac{Mass}{Volume}

\\ \tt\longmapsto Density=\dfrac{2.22}{0.08}=27.75g/cm^3

7 0
2 years ago
How would poisoning proton pumps impact anion uptake? see section 36.3 ( page 755) ?
boyakko [2]
The answer to this question would be: the anion uptake will be decreased

If proton pump function is decreased, that means the amount of proton that goes outside the cell will be decreased. It will cause the anion uptake decreased too because some anion needs to go inside cells by co transporting with the proton.
7 0
4 years ago
List three organelles found in cells and describe their functions.
kozerog [31]

nucleus: controls cell

cell membrane: controls what goes in and out of a cell

cytoplasm: where chemical reactions take place

3 0
3 years ago
Read 2 more answers
1. When 50.0 mL of water at 80.0°C was mixed with 50.0 mL of water in a calorimeter at
ddd [48]

Answer:

The heat capacity of the calorimeter is 5.11 J/g°C

Explanation:

Step 1: Given data

50.0 mL of water with temperature of 80.0 °C

Specific heat capacity of water = 4.184 J/g°C

Consider the density of water = 1g/mL

50.0 mL of water in a calorimeter at 20.0 °C

Final temperature = 47.0 °C

Step 2: Calculate specific heat capacity of the water in calorimeter

Q = Q(cal) + Q(water)

Q(cal) = mass * C(cal) * ΔT

Qwater = mass * Cwater * ΔT

Qcal = -Qwater

mass(cal) * C(cal) * ΔT(cal) =  mass(water) * C(water) * ΔT(water)

50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)

1350 * C(cal) = 6903.6

C(cal) = 5.11 J/g°C

The heat capacity of the calorimeter is 5.11 J/g°C

8 0
3 years ago
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