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LekaFEV [45]
4 years ago
9

Question 1

Chemistry
1 answer:
luda_lava [24]4 years ago
4 0

Answer : The name of the given hydrocarbon is, pentane.

Explanation :

The basic rules for naming of organic compounds are :

  • First select the longest possible carbon chain.
  • For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
  • A suffix '-ane' is added at the end of the name of alkane.
  • If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

In the given hydrocarbon, the longest possible carbon chain number is 5 that means we add prefix 'pent' and suffix '-ane'. So, the name of given hydrocarbon will be, pentane.

Hence, the correct option is, pentane.

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The following reaction shows the products when sulfuric acid and aluminum hydroxide react
ivann1987 [24]

Answer:

I can't see the picture so I'm not able to help you.

Explanation:

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5 0
3 years ago
What mass of water at 10.0 oC would be required to cool 50.0g of a metal having a specific heat of
Stels [109]
<h3>Answer:</h3>

The mass of water required is 50.2 g

<h3>Explanation:</h3>
  • Quantity of heat of a substance is calculated by multiplying the mass of a substance by the specific heat capacity and the change in temperature.

That is;

Q=m×c×ΔT

In this case, water is used to cool a metal, therefore, water will gain heat while the metal will lose heat.

Heat gained is equivalent to heat lost

Heat gained by water = Heat lost by the metal

Step 1: Heat lost by the metal

Mass of the metal  = 50.0 g

Specific heat capacity of metal = 0.60 J/g°C

Change in temperature (ΔT) = 70°C

Heat lost by the metal = 50 g× 0.6 × 70

                                     = 2100 Joules

Step 2: Heat gained by water

Mass of water = x g

Specific heat capacity of water = 4.18 J/g°C

Temperature change = 10 °C

Heat gained by water = x g × 4.18 × 10

                                    = 41.8x joules

Step 3: Mass of water

Heat gained by water = heat lost by the metal

41.8x = 2100

   x = 50.239 g

      = 50.2 g (1 d.p)

The mass of water is 50.2 g

7 0
3 years ago
A gas stream contains 18.0 mol% hexane and the remainder nitrogen. the stream flows to a condenser, where its temperature is red
SVEN [57.7K]
There is no question specified for this problem. Based on my understanding, there are 5 possible questions that would come up. Once you know them, the whole problem is solved. They are:

a.) The amount of entering gas stream (denoted as A)
b.) The amount of exiting gas stream (denoted as B)
c.) The amount of condensate (denoted as C)
d.) Mole fraction of liquid hexane (x,h)
e.) Mole fraction of liquid nitrogen (x,n)

In order to solve this, we apply mass balance.
Overall Mass Balance: A = B + C
Hexane Balance: 0.18A = 0.05B + 1.5
Nitrogen Balance: 0.82A = 0.95B + (C - 1.50)

Solving the equations simultaneously:
0.18(B+C) = 0.05B + 1.50
C = (1.5 - 0.13B)/0.18

0.82[B + ((1.5 - 0.13B)/0.18)] = 0.95 + ((1.5 - 0.13B)/0.18) - 1.5
Solving for B,
B = 1 L/min
Then,
C = (1.5 - 0.13(1))/0.18 = 7.611 L/min
A = 1 L/min + 7.611 L/min = 8.611 L/min
x,h = 1.5/C = 1.5/7.611 = 0.197
x,n = 1 - 0.197 = 0.803

As a summary, the possible answers to the possible questions are:
a.) 8.611 L/min entering gas stream
b.) 1 L/min exiting gas stream
c.) 7.611 L/min condensate
d.) The condensate is 19.7 mol% hexane
e.) The condensate is 80.3 mol% nitrogen

4 0
3 years ago
Keera decided to investigate how the growth of plants is affected by the time of day it is watered. She placed an equal amount o
Arada [10]
The time of day the plants are watered is the independent variable. Keera did everything else the same for each plant.
7 0
3 years ago
1) A container holds 5,000 mL of CO2 at 742 atm. What will be the volume of the CO2 if the pressure is increased to 795 atm?​
Paladinen [302]

Answer:

V_2=4667mL

Explanation:

Hello there!

In this case, based on the given information, it is possible for us to provide an answer to this question by using the Boyle's law as directly proportional relationship between volume and pressure:

P_2V_2=P_1V_1

Thus, by solving for the final volume, V2, as required, we obtain the following:

V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{5,000mL*742atm}{795atm}\\\\V_2=4667mL

Regards!

7 0
3 years ago
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