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kykrilka [37]
2 years ago
9

A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

icroliters must be administered?
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many microliters must be administered?
Chemistry
1 answer:
oksian1 [2.3K]2 years ago
4 0

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

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Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of
Andrej [43]

Given the mass of hydrogen = 16.0 oz

Converting 16.0 oz hydrogen to pounds (lb) using the conversion factor 1 lb = 16 oz:

16.0 oz * \frac{1 lb}{16 oz} =1 lb

Converting 16.0 lb to g using the conversion factors 1 kg = 2.2 lb, 1 kg = 1000 g:

1lb * \frac{1kg}{2.2lb}*\frac{1000g}{1kg}= 454.5 g

Heat of combustion of hydrogen = 142 J/g

Calculating the heat released when 16.0 oz is combusted:

454.5g H_{2} * \frac{142 J}{g} *\frac{1 kJ}{1000J}=64.5kJ


5 0
2 years ago
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Franklin was performing an experiment by combining hydrochloric acid and sodium hydroxide. He measured the mass of his reactant
VARVARA [1.3K]

Answer:

B

Explanation:

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1 year ago
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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
2 years ago
A lab director asks two students, Lydia and Damien, to each select a bottle of a concentrated weak base. When they reach the
Hunter-Best [27]

Answer:

First one is 5.0 M ammonia and the Second one ?

Explanation:

6 0
2 years ago
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A molecule contains 23.24 g iodine (I),
gavmur [86]
Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
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