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2 years ago

A strand of DNA has the following string of bases:

Chemistry

1 answer:

Ganezh [65]2 years ago

5 0

Answer:

ok so first of all try working hard then ill answer u

Explanation:

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How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at −57.1 °c. give your answer as a po

kolezko [41]

To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g;
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348
= 26250.0348 J or 26.250 kJ

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3 years ago

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2 years ago

Both of the silicon-chlorine single bonds in sicl4 are polar. in which direction should the polarity arrows point?

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3 years ago

27. A solution has a pH of2. Which best describes the solution? (a point)

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Explanation:

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3 years ago

Read 2 more answers

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.

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3 years ago