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Natalija [7]
3 years ago
14

.............................................................. ......

Chemistry
1 answer:
ollegr [7]3 years ago
5 0

Answer:

the question is blank. what's up?

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Lipids is a kind of biochemical that does not dissolve in water and makes up the cell walls fats oil and waxes
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The ph of a solution for which [OH]= 1.0 x 10^-4 is?<br><br> A 10<br> B 14<br> C 11<br> D 9<br> E 7
zhenek [66]
To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

pH= -log (1.0x10-10)= 10

answer is A
8 0
3 years ago
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
pochemuha

Complete question:

Write the condensed formula from left to right, starting with (CH3)x where x is a number.

See attached image for the structure formula of the compound

Answer:

(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane

Explanation:

If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.

Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).

The condensed formula will be written as;

(CH₃)₂CHC(CH₃)₃

This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane

3 0
3 years ago
Explain what helps convert non-rich oxygen blood cells back to oxygen rich cells?
Reil [10]

Answer:

As blood travels through the body, oxygen is used up, and the blood becomes oxygen poor. Oxygen-poor blood returns from the body to the heart through the superior vena cava (SVC) and inferior vena cava (IVC), the two main veins that bring blood back to the heart.

Explanation:

3 0
2 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
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