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Oduvanchick [21]
2 years ago
6

Which of the following are examples of types of mixtures?

Chemistry
1 answer:
aev [14]2 years ago
6 0
I think it would be homogeneous and heterogeneous
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Categorize each hydrocarbon as being saturated or unsaturated.
Black_prince [1.1K]

As you have not provided the options, still we can figure out the answer by understanding the key difference between saturated and unsaturated hydrocarbons.

SATURATED HYDROCARBONS are those hydrocarbons which only consist of a carbon carbon single bonds. All the bonds are sigma there are no pi bonds at all. Examples are shown below.

While, UNSATURATED HYDROCARBONS are those hydrocarbons which may contain either a double bond or triple bonds or both of them between the carbon atoms as shown below.

6 0
3 years ago
What items are true about a block of ice at -10°C as you continue to apply heat
zhenek [66]

1. its temperature will rise continuously until it melts

I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting

5 0
3 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
1. Review the information in the table below. Use the information to calculate rate of how many sandwiches are made in a 10-minu
PilotLPTM [1.2K]

Answer:

The answer to your question is given below.

Explanation:

Rate is simply defined as quantity per unit time. Mathematically it is represented as:

Rate = Quantity /time

Thus, we can obtain the rate as follow:

1. Quantity = 20 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 20/10

Rate = 2 sandwich per mins

2. Quantity = 30 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 30/10

Rate = 3 sandwich per mins

3. Quantity = 40 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 40/10

Rate = 4 sandwich per mins

4. Quantity = 50 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 50/10

Rate = 5 sandwich per mins

Thus, the complete table is given as follow:

Quantity >> Unit of measure >> Rate

10 >>>>>>> 10 min >>>>>>>>>> 1

20 >>>>>>> 10 min >>>>>>>>>> 2

30 >>>>>>> 10 min >>>>>>>>>> 3

40 >>>>>>> 10 min >>>>>>>>>> 4

50 >>>>>>> 10 min >>>>>>>>>> 5

4 0
2 years ago
Given that you added about 5ug of purified acid phosphatase to tube A, calculate the amount of acid phosphatase that was present
Reil [10]

Hey there!:

Amount of purified acid phosphatase added to tube A = 5 ug ( micrograms )

Amount of acid phosphatase present in 400 ug of wheat germ extract  in tube B :

1 / 100 = 5x / 400 =

100 x = ( 0.5 ) ( 400 ) =

x = (0.5 ) ( 400 ) = 100

x = 200 / 100

x =  2 ug


The amount of acid phosphatase present in 400 ug of the  wheat germ in tube B is 2 ug


Hope that helps!

6 0
3 years ago
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