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Taya2010 [7]
3 years ago
14

3) Convection can only happen in what? *

Chemistry
1 answer:
otez555 [7]3 years ago
3 0

Answer:

Convection can only happen with liquids and gases.

Hope it helps :)

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Alja [10]
Producers produce a product
Consumers buy that product
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3 years ago
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Why do acids and bases neutralize eachother?
timofeeve [1]

Answer:

Explanation:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

5 0
3 years ago
If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a
strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

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 ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

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

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

3 0
3 years ago
Is alcohol a pure substance or a mixture
Genrish500 [490]
Alcohol is a pure substance
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3 years ago
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A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

The volume of the gas after it is heated.

Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L

Hence, this is the required solution.

3 0
3 years ago
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