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Anastaziya [24]
3 years ago
15

What is meant by enthalpy change?

Chemistry
1 answer:
Vinil7 [7]3 years ago
7 0
Enthalpy change refers to the overall amount of heat added or lost with each step as you progress through your reaction.
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What is the volume of a liquid that has a density 3.65g/ml and a mass of 5.61g
Lorico [155]

Answer:

<h2>1.54 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{5.61}{3.65}  \\  =1.536986...

We have the final answer as

<h3>1.54 mL</h3>

Hope this helps you

3 0
3 years ago
The atomic mass of carbon -13 is 13. It has six protons. How many neutrons does the isotope have​
marishachu [46]

Answer:

13 - 6 =7

Explanation:

8 0
3 years ago
If a reaction taking place in a flask shows a drop in temperature, which type of reaction is it? Options: exothermic reactions e
liubo4ka [24]

Answer:

exothermic reaction

Explanation:

If there is a drop in temperature, then energy was lost to the surroundings because temperature is the average measure of kinetic energy. An exothermic reaction would result in this lost of energy. An endothermic reaction would absorb energy and make the temperature rise.

4 0
3 years ago
Read 2 more answers
56789x09876<br><br><br><br><br><br><br> Does anyone want to rp?? Any kind of rp I'm just bored
Alja [10]
The answer would be 560,848,164 because the 0 at the start WOULDNT matter
7 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
3 years ago
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