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Usimov [2.4K]
3 years ago
14

“While observing a chemical reaction, how can you tell which reactant is limiting?”

Chemistry
1 answer:
bulgar [2K]3 years ago
3 0

Answer:

Then the substance would be the first to run out which signifies the completing of the reaction.

Explanation:

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Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?
8_murik_8 [283]
E = mgh

E = 70*9.81*80

E = 54936 J
5 0
3 years ago
What type of nuclear decay is shown by the reaction below?
pantera1 [17]

Answer:

B. Alpha

Explanation:

Alpha decays always split into an element and He.

Beta decays always split into an element and e- (eletrons).

Gamma decays always split into an element and radiation.

3 0
3 years ago
If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

3 0
2 years ago
Write the chemical equations for the neutralization reactions that occurred when hcl and naoh were added to the buffer solution.
irga5000 [103]
A base and an Acid always react to form a salt and water

So, HCl + NaOH —> NaCl + HOH
5 0
3 years ago
0.820 L has a mass of 2.56kg what is the density of this liquid in kg/L
3241004551 [841]

Density ρ=\frac{m}{v}

=\frac{2.56 kg}{0.820 L} \\=3.121kg/L

8 0
3 years ago
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