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zimovet [89]
3 years ago
10

Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurate

ly prepared in a single step due to the small amount of mass that would need to be measured. Therefore, you decided to conduct this as a two step process. You will create a high concentration stock solution in one 10 mL flask. You then transfer 0.1 mL into a second 25 mL volumetric flask that will be the concentration of 6 ng/μL. What is the mass of chemical you need to add to the stock flask to achieve the desired concentration after dilution?
Chemistry
1 answer:
Over [174]3 years ago
4 0

Answer:

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25)/ 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg

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Explanation:

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3 years ago
Which is an example of a bioresource used to build a home?
lutik1710 [3]
The answer should be A.
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4 0
3 years ago
how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin
STALIN [3.7K]

Answer:

1. q.

2. 2q.

3. 3q.

4. 6q.

Explanation:

We'll begin by calculating the specific heat capacity of the liquid. This can be obtained as follow:

Mass (m) = 25 g

Change in temperature (ΔT) = 20 °C

Heat (Q) = q

Specific heat capacity (C) =?

Q = MCΔT

q = 25 × C × 20

q = 500C

Divide both side by 500

C = q/500

C = 2×10¯³ qg°C

Therefore, the specific heat capacity of liquid is 2×10¯³ qg°C

Now, we shall determine the heat required to produce the various change in temperature as follow:

2. Mass (m) = 50 g

Change in temperature (ΔT) = 20 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 20

Q = 2q.

Therefore, the heat required is 2q.

3. Mass (m) = 25 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 25 × 2×10¯³ × 60

Q = 3q.

Therefore, the heat required is 3q.

4. Mass (m) = 50 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 60

Q = 6q.

Therefore, the heat required is 6q.

4 0
3 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
3 years ago
three moles of sodium carbonate are mixed with two moles of lead nitrate in aqueous solution, leading to formation of a solid pr
aleksandrvk [35]

There are 4 moles of spectator ions that remain in solution.

The equation of the reaction is;

Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)

We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.

For Na2CO3

1 mole of Na2CO3 yields 2 moles of NaNO3

3 moles of Na2CO3 yields  3 × 2/1 = 6 moles of NaNO3

For Pb(NO3)2

1 mole of Pb(NO3)2 yields 2 moles of NaNO3

2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3

We can see that Pb(NO3)2 is the limiting reactant.

Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.

Learn more: brainly.com/question/22885959

6 0
3 years ago
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