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natulia [17]
3 years ago
8

For lithium isotopes, 92% are Lithium-6 and 7.6% are Lithium-5. How many neutrons am I most likely to find in a lithium sample?

Biology
1 answer:
torisob [31]3 years ago
7 0
Naturally occurring lithium<span> (chemical symbol Li) (relative </span>atomic mass<span>: 6.941(2)) is composed of two stable </span>isotopes<span>, </span>lithium-6<span> and </span>lithium<span>-7, with the latter being far </span>more<span> ... </span>Lithium-4 contains threeprotons<span> and one </span>neutron<span>. ... An accurate </span>atomic mass<span> for </span>samples<span> of </span>lithium<span> cannot be measured for all sources of </span>lithium<span>.</span>
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Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

7 0
3 years ago
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