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4 years ago

Does anyone know the answer to this chemistry question?!?

Chemistry

1 answer:

Pepsi [2]4 years ago

7 0

Answer:

Mass of HCl = 73 g

Explanation:

Given data:

Mass of hydrogen = 2 g

Mass of HCl = ?

Solution:

First of all we will write the balance chemical equation:

H₂ + Cl₂ → 2HCl

Number of moles of hydrogen = 2 g/ 2g/mol

Number of moles of hydrogen = 1 mol

Form balanced chemical equation compare the moles hydrogen with HCl.

H₂ : HCl

1 : 2

Mass of HCl:

Mass of HCl = number of moles × molar mass

Mass of HCl = 2 mol × 36.5 g/mol

Mass of HCl = 73 g

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Select the correct answer. Which substance is made of polymers? A. marble B. protein C. salt D. steel

Greeley [361]

Answer:

the answer is B proteins

Explanation:

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3 years ago

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A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0

3 years ago

How many grams of C3H8 are in a 7 L tank at 293 K and 5.45 atm?

Sophie [7]

1.59 moles

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8 0

3 years ago

Write the overall reaction for cellular respiration

goblinko [34]

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

7 0

3 years ago

Could someone explain how they got this answer, explain step by step plz

gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

7 0

4 years ago