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alex41 [277]
3 years ago
12

18)

Chemistry
1 answer:
solmaris [256]3 years ago
8 0
The answer is option D, limiting reactant.

Oftentimes in chemistry, there are leftovers of certain chemicals in reactions. When not all of a certain reactant is used up in a reaction, it is known as the excess reactant (since there is an excess of it). The reactant that's completely used up is known as the limiting reactant (since it limits the reaction from going any further)

-T.B.
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Fossil originally has 254 g of carbon-14 but currently has 127 g of carbon-14 the fossil is most likely
KATRIN_1 [288]

Answer:

The answer to your question is: letter C) 5730

Explanation:

Data

Originally Carbon - 14   254 g

Currently Carbon - 14   127 g

Half-life of Carbon -14 is when half of the original amount of Carbon disintegrates.

Then, if originally there were 254 and now there are 127, only one half life passed.

Half life of carbon -14 is 5730 years old.

8 0
4 years ago
If a student pushes a book across a table with a force of 6 N and the force of friction is 4 N. What is the net Force acting on
Sedaia [141]

Answer:

The net force is 2 N in the direction the book is being pushed.

Explanation:

6 0
3 years ago
Read 2 more answers
If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the wate
natka813 [3]

Answer:

T2 = 29°C

Explanation:

Given data:

Heat added = 420 j

Mass of water = 25 g

Initial temperature = 25°C

Final temperature = ?

Solution;

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values.

420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)

420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)

420 j = 104.5  j/°C × (T2 - 25°C)

420 j /104.5  j/°C = T2 - 25°C

4°C + 25°C = T2

T2 = 29°C

4 0
4 years ago
9
yaroslaw [1]

Answer:

T2 = 36.38°C

Explanation:

Given data:

Mass of water = 75 g

Initial temperature = 30 °C

Final temperature = ?

Heat absorbed = 2000 J

Solution:

Specific heat capacity of water is 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

2000 J = 75 g×4.18 J/g.°C × T2- T1

2000 J = 313.5 J/°C × T2- T1

2000 J = 313.5 J/°C × T2 - 30 °C

2000 J / 313.5 J/°C  = T2 - 30 °C

6.38 °C = T2 - 30 °C

T2 = 6.38 °C +  30°C

T2 = 36.38°C

7 0
3 years ago
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

4 0
3 years ago
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