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vladimir1956 [14]
3 years ago
5

Chem please help theres a photo attatched

Chemistry
2 answers:
a_sh-v [17]3 years ago
8 0

Answer:

the top one is 100.000 and bottom one is 100.000

Explanation:

pashok25 [27]3 years ago
4 0
The first one is 100.000% and the second one is 97.24%
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A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill
tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

N_{t} = N_{0}e^{-\lambda t}

Where:

λ: is the decay constant = ln(2)/t_{1/2}

t_{1/2}: is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg

Now, at 7:00 pm we have:

t = 7:00 pm - (5:00 pm + 90 min) = 30 min

N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg    

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0
3 years ago
Which of the following is not equal to 120 centimeters?
tia_tia [17]
<h3>Answer:</h3>

0.012 dekameters (dkm)

<h3>Explanation:</h3>

<u>We are given;</u>

  • 120 centimeters( cm).

Required to identify the measurements that is not equivalent to 120 cm.

  • Centimeters are units that are used to measure length together with other units such as kilometers(km), meters (m), millimeters (mm), dekameters (dkm), etc.
  • These units can be inter-converted to one another using suitable conversion factors.
  • To do this, we are going to have a table showing the suitable conversion factor from one unit to another.

Kilometer (km)

10

Decimeter (Dm)

10

Hectometer (Hm)\

10

Meter (m)

10

Dekameter (dkm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore;

To convert cm to km

Conversion factor is 10^5 cm/km

Thus;

120 cm = 120 cm ÷ 10^5 cm/km

            = 0.0012 km

To convert cm to dkm

Conversion factor is 10 cm/dkm

Therefore,

120 cm = 120 cm ÷ 10 cm/dkm

            = 12 dkm

To convert cm to m

The suitable conversion factor is 10^2 cm/m

Thus,

120 cm = 120 cm ÷ 10^2 cm/m

           = 1.2 m

To convert cm to mm

Suitable conversion factor is 10 mm/cm

Therefore;

120 cm = 120 cm × 10 mm/cm

            = 1200 mm

Therefore, the measurement that is not equal to 120 cm is 0.012 dkm

4 0
3 years ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
An earthquake’s magnitude is a measure of the
marissa [1.9K]
An earthquake's magnitude is a measure of how much energy an earthquake releases. Typically, the richter scale is used.
8 0
3 years ago
Read 2 more answers
Organic acids and bases in their ___________ form are soluble in organic solvents, but the corresponding _________ are more wate
gizmo_the_mogwai [7]

Answer:

a. neutral

b. salts

c. salt

Explanation:

Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.

Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again

RA + H + B => RA + HB

H belongs to A, but it is written as it is involved in the neutralization reaction.

On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.

When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH

3 0
3 years ago
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