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3 years ago

Describe how to prepare 10 ml of 5, 10, 15, and 20 micro M CV solution using a 25 microM CV stock solution

1 answer:

4 0

A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a stock solution whose concentration is known.

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The amount of water vapor in the air, compared to the maximum amount that can be held at a specific air temperature is called:

Lisa [10]

Answer: D relative humidity

Explanation: Just took the test.

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3 years ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

3 years ago

How many kilometers are there in 4.5 x 10^3 meters ?

Len [333]

1 meter = .001 kilometers so 4.5 x 10^3 = 4500 so .001 x 4500 = 4.5 kilo

7 0

3 years ago

What is the formula name for Co3N2

Ede4ka [16]

Answer:

Cobaltous Nitride,I think so anyway.......

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3 years ago

The density of benzene at 15 ∘C is 0.8787 g/mL. Calculate the mass of 0.1200 L of benzene at this temperature.

Pavel [41]

Use the density to convert volume into mass.

since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.

0.1200 Liters= 120.0 mL

120.0 mL (0.8787 grams/ 1 mL)= 105 grams

3 0

3 years ago

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