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myrzilka [38]
2 years ago
5

Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui

nine? (Atomic weight: N = 14.01).
Chemistry
1 answer:
Furkat [3]2 years ago
6 0

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

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Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.
kupik [55]
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,

                    2H2 + O2 --> 2H2O

The masses of each of the reactants are calculated below.

          2H2 = 4(1.01 g) = 4.04 g
          O2 = 2(16 g) = 32 g

Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
        (1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2

Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.


<em>Answer: 1.22 g of oxygen</em>
4 0
2 years ago
The partial pressures in an equilibrium mixture of NO, Cl2, and NOCI at 500 K are as follows: PNo = 0.240 atm, Pel2 = 0.608 atm,
maxonik [38]

Answer : The value of K_p at temperature 500 K is 52.0

Explanation : Given,

Partial pressure of NO at equilibrium = 0.240 atm

Partial pressure of Cl_2 at equilibrium = 0.608 atm

Partial pressure of NOCl at equilibrium = 1.35 atm

The balanced equilibrium reaction is,

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(1.35)^2}{(0.240)^2(0.608)}

K_p=52.0

Therefore, the value of K_p at temperature 500 K is 52.0

3 0
2 years ago
The table shows the amount of radioactive element remaining in a sample over a period of time.
MrRissso [65]

Answer:

8,000 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

Part 1: What is the half-life of the element? Explain how you determined this.

  • The half-life of the element is 1,600 years.

Half-life time is the time needed for the reactants to be in its half concentration.

The sample stats with 56.0 g and reaches its half concentration (28.0 g) after 1,600 years.

<em>So, the half-life of the sample is 1,600 years.</em>

<em></em>

Part 2: How long would it take 312 g of the sample to decay to 9.75 grams? Show your work or explain your answer.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).

[A] is the remaining concentration of the sample ([A] = 9.75 g).

<em>∴ t = (1/k) ln([A₀]/[A])</em> = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = <em>8,000 years</em>.

6 0
2 years ago
Write the lewis structure for ch2clcoo−. assign a formal charge for any atom with a non-zero formal charge.
Murljashka [212]
The Lewis structure of Chloroacetate (H₂CClCO₂) is given below. In structure it is shown that carbon has a double bond with one oxygen atom and two single bonds with CH₂Cl and O⁻.

Formal Charge;
                        Formal charge is caculated as,

Formal charge  =  # of valence e⁻ - [# of lone pair of e⁻ + 1/2 # of bonded e⁻]

Formal charge on Oxygen (Highlighted Red);

Formal charge  =  6 - [ 6 + 2/2]

Formal charge  =  6 - [6 + 1]

Formal charge  =  6 - [7]

Formal charge  =  -1

8 0
2 years ago
How many grams of caf2 would be needed to produce 8.41×10-1 moles of f2?
geniusboy [140]
Answer: 65.7 grams

Explanation:

1) ratio

Since 1 mole of CaF2 contains 1 mol of F2, the ratio is:

1 mol CaF2 : 1 mol F2

2) So, to produce 8.41 * 10^ -1` mol of F2 you need the same number of moles of CaF2.

3) use the formula:

mass in grams = molar mass * number of moles

molar mass of CaF2 = 40.1 g/mol + 2 * 19.0 g/mol = 78.1 g/mol

mass in grams = 78.1 g/mol * 8.41 * 10^ -1 mol = 65.7 grams
5 0
2 years ago
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