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Crazy boy [7]
3 years ago
9

Hello friends good afternoon is anybody here​

Chemistry
1 answer:
rewona [7]3 years ago
3 0
Hello friend :)!!!!!!!!!
You might be interested in
Can someone help me here please i need this :(​
podryga [215]
Start by adding the numbers then divide
7 0
2 years ago
What are some examples of supersaturated solutions? Give 10.​
Phantasy [73]

Examples include carbonated water (i.e. soda water); honey; sugar syrup (used in confectionery); supersaturated drug delivery systems. "SDDS"; and sodium acetate solutions prepared from 160 g NaOAc and 30 mL water.

7 0
3 years ago
WILL GIVE POINTS/MARK BRAINLIST! Question 6: Types of Molecular Shapes (3 points) Identify the following molecular shapes. (0.5
Vanyuwa [196]

Answer:

A. tetrahedral

B. angular or bent

C. angular

D. trigonal planar

E. linear

F. linear

Explanation:

8 0
2 years ago
Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are
pantera1 [17]

Answer:

1. Options A and B

2. Options B and C

3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol

Explanation:

1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.

Therefore, since reaction A and B have ∆G to be positive, they are unfavorable

2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.

Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.

3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol

B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol

4 0
2 years ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
3 years ago
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