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3 years ago

Names of nitrogenousbonucleic bases in dna

Chemistry

1 answer:

Crazy boy [7]3 years ago

5 0

Explanation:

The four nitrogenous bases present in DNA are adenine (A), guanine (G), cytosine (C) and thymine (T).

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A spherical balloon initially contains 25m3 of helium gas at 20o C and 150 kPa. A valve is now opened and the helium is allowed

Tpy6a [65]

Find the attachment for complete solution

6 0

2 years ago

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0

3 years ago

An aqueous solution of licl is 34.0 % licl. what mass of water is present in 250.0 g of the solution? (density of water is 1.00

adell [148]

<span>If the aqueous solution is 34% Licl then it is 100 - 34% water = 66% From the calculation we've found out that it is 66% water. Then we need to find the weight from a 250 g solution. 66/100 * 250 = 165g Hence it is 165g</span>

4 0

3 years ago

Read 2 more answers

If the amount of Adams if each type on the left and right sides of a reaction difference what must be done to balance it

Ivanshal [37]

Hey there,

Your answer would be

Coefficients are placed in front of the reactants and/or products

Hope this helps,

<h2>- <em>Mr. Helpful</em></h2>

8 0

3 years ago