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andreyandreev [35.5K]

3 years ago

14

Which of the following best describes a single replacement reaction?

1 answer:

Lady_Fox [76]3 years ago

6 0

Answer:

One element takes the place of another in a compound

Explanation:

I just took a test for it and got it right. :)

Hope This Helps :)

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OHC-CH2-CH2-CH2-CHO<br><br> What is the IUPAC name of this compound ?

Rufina [12.5K]

Answer:

Pentan_1,5_di-al

Explanation:

OHC-CH₂-CH₂-CH₂-CHO

This is Pentan_1,5_di-al

If we break this compound, we will observe that there is presence aldehyde group and hence the functional group "al". This aldehyde is bonded to carbon 1 and carbon 5 respectively.

Also the pentan is due to presence of 5 carbon atoms.

Therefore, the IUPAC name of this compound (OHC-CH₂-CH₂-CH₂-CHO) is Pentan_1,5_di-al

5 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

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

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

3 years ago

What type of energy is present in a barbell being lifted or a shot-put being thrown?

denpristay [2]

The type of energy used is kinetic energy. Kinetic energy is the energy of motion.

8 0

3 years ago

Read 2 more answers

Transpiration is the process by which ____.<br>

saul85 [17]

Answer:

Plants add water to the atmosphere

Explanation:

3 0

3 years ago

Read 2 more answers

Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce

kkurt [141]

Explanation:

The given data is as follows.

P = 3 atm

= $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$

= $3.03975 \times 10^{5} Pa$

$V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$ (as 1 L = 0.001 $m^{3}$ ),

$V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

W = $P \times \Delta V$

or, W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

W = $P \times (V_{2} - V_{1})$

= $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

= 1823.85 Nm

or, = 1823.85 J

As internal energy of the gas $\Delta E$ is as follows.

$\Delta E$ = Q - W

= 800 J - 1823.85 J

= -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0

3 years ago