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mafiozo [28]
1 year ago
14

an effective hot pack salt can release 5850 j of heat to the surroundings. how many grams of licl are needed to release this amo

unt of heat?
Chemistry
1 answer:
Montano1993 [528]1 year ago
6 0

4.01 grams of Licl are needed to release this amount of heat

<h3>What does one gram weigh?</h3>

The gram is a unit of mass in the International System of Units (SI) that is equal to one thousandth of a kilogram. It was originally known as the gramme. Gram. This pen cap weighs around one gram. A weight scale like this one may provide a precise mass readout for many different things.

Mass of Licl required to release 5850J of heat to the surroundings.

Let x g Licl required.

Then x g ×∆H = 5850J

x g × 1.46 × 10³ = 5850

x = 5850/ 1.46 × 10³

x = 4.0068

so 4.01g Licl required.

To know more about gram visit:

brainly.com/question/12127497

#SPJ4

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The black fur absorbs all the colors where the white fun reflects all the colors because we see the green grass it absorbs all the colors except green and the yellow dandelion is reflecting yellow and orange because it’s color is like a mix of them but it still absorbs the other colors
4 0
3 years ago
Read 2 more answers
The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of
Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

C

.

Water has a specific heat of approximately  

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

C

, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

times as much heat to increase its temperature by  

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

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

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

5 0
3 years ago
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
3 years ago
Definition: This is a net gain or loss of electrons.
AlekseyPX

Answer:

Elecric charge/ Electricity  

Explanation:

Electric charge is the net gain or loss of electrons

I hope im right!!

3 0
3 years ago
What is the majority of 0.50 g of na dissolved in a 1.5 L solution
Oxana [17]

The correct answer is 0.014467 M.  

Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.  

The molar mass of sodium (Na) is 23.0 g/mol

Thus, 1 mole of Na contains 23.0 g

Now, x moles of Na contains 0.50 g

Moles = 0.50 × 1 / 23.0

Moles = 0.50 / 23.0  

= 0.0217 moles of Na

Molarity = Number of moles / liters of solution

= 0.0217 / 1.5

= 0.014467 M


8 0
3 years ago
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