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Arte-miy333 [17]
3 years ago
8

The term that describes two or more objects in a system reaching the same temperature is

Chemistry
1 answer:
Galina-37 [17]3 years ago
7 0

The correct answer is Thermal Equilibrium

Explanation:

The term "thermal equilibrium" is used when two or more objects have the same temperature and therefore there is not an exchange of heat between them. This occurs when the objects had a different temperature at the beginning but due to a close contact heat is transferred from one object to the other until an equilibrium or same temperature is reached. For example, a hot cup over a table or any other surface will transfer the heat to the surface, but after some time both the cup and the surface will have the same temperature or will reach thermal equilibrium.

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Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma
Lina20 [59]

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

-CH_2=CH-CH=CH2-

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D:  A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

8 0
3 years ago
How much (Q) heat is needed to melt 35 g of iodine? Hf = 61.7 J/g.
aev [14]

Taking into account the definition of calorimetry and latent heat, a heat of 2159.5 J is needed to melt 35 g of iodine.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

<h3>Heat needed to melt iodine</h3>

In this case, you know:

  • m= 35 g
  • L=61.7 \frac{J}{g}

Replacing in the definition of latent heat:

Q= 35 g× 61.7 \frac{J}{g}

Solving:

<u><em>Q=2159.5 J</em></u>

Finally, a heat of 2159.5 J is needed to melt 35 g of iodine.

Learn more about calorimetry:

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#SPJ1

8 0
2 years ago
The following equation represents the type of reaction called
Eva8 [605]

Answer: C REDUCTION

Explanation:

Guessed after knowing oxidation isn't the answer. Got right

7 0
3 years ago
0.32 g of a walnut is burned under an aluminum can filled with 58.1 mL of water. The water temperature in the can increases by 3
balu736 [363]

Answer:

1.8 × 10² cal

Explanation:

When 0.32 g of a walnut is burned, the heat released is absorbed by water and used to raise its temperature. We can calculate this heat (Q) using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity of water

m: mass of water

ΔT: change in the temperature

Considering the density of water is 1 g/mL, 58.1 mL = 58.1 g.

Q = c × m × ΔT

Q = (1 cal/g.°C) × 58.1 g × 3.1°C

Q = 1.8 × 10² cal

3 0
3 years ago
How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac
bogdanovich [222]
You need the set of reactions that goes from ammonia to nitric acid.
<span>
1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

2) 2NO(g)+O2(g)-->2NO2(g)

3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)

State the ratio of moles of HNO3 to NH3:

4 moles of NH3 produce 4 mole of NO,

4 moles of NO produce 4 moles of NO2

4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

=> (8/3) moles HNO3 : 4 moles NH3

Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3

Use proportions:

(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

Convert moles to grams:

molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

Answer: 3213 g.
</span>
3 0
3 years ago
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