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Nady [450]
2 years ago
15

QUICK CHECK

Chemistry
1 answer:
mamaluj [8]2 years ago
4 0

The structure that corresponds to methylpropylether is the structure CH3 - O - C3H7 in option C.

<h3>What is structure?</h3>

The structure of a compound is defined as the way in which the atoms in the compound are arranged. We know that a compound is composed of atoms and these atoms are arranged in particular patterns in space. This is the stereochemistry of the molecule.

Now we know that methylpropylether is a compound that contains the methyl group and the propyl group separated by a oxygen atom as is typical of all ethers.

As such, the structure that corresponds to methylpropylether is the structure CH3 - O - C3H7 in option C.

Learn more about methylpropylether:brainly.com/question/28047849

#SPJ1

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Triss [41]

Answer: (C) Dissolution is the process by which a solute form a solution in a solvent.

Explanation: Dissolution can be described as the process by which a solute dissolves in a solvent to give a solvent.

The solute may be a solid, liquid or gas and solvent is usually a liquid but can also be liquid, gas or supercritical fluid. The most common solvent is water and it is referred to as the universal solvent.

Example are:

1.Stirring of sugar in water.

In this case the solute is the sugar and the solvent is water.

2.Organic substances dissolves readily in solvent like kerosene and benzene and they are known as non-polar solvent while the organic substances are the solute

4 0
3 years ago
How many liters in 4.4 grams of CO2 at STP?
d1i1m1o1n [39]

Answer:

2.24 Liters are in 4.4 grams of CO2 at STP

5 0
4 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come
Elden [556K]

Answer:

B

Explanation:

For solving this we need a heat balance

Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}

By changing the corresponding relations, we have

m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\

By cancelling similar factor, we obtain

\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\

Which means that the change of temperature in A is twice the change of B

3 0
3 years ago
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Answer:

Explanation:

The triple point of carbon dioxide is 5.11 atmosphere at -56.6 degree celsius . At pressure greater than 5.11 , solid carbon dioxide liquefies ,  as it is warmed. At pressure lesser than 5.11 atmosphere , it will go into gaseous state without liquefying . Excessive pressure helps liquification process.

So maximum pressure required is 5.11 atmosphere. Beyond this pressure , solid CO2 will liquify.

3 0
3 years ago
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