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vodka [1.7K]
3 years ago
12

You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that

contains 5% naocl by weight. assuming that the density of the stock solution is 1.0 g/ml, how many milliliters of bleach should you add to each liter of test solution?
Chemistry
1 answer:
Alecsey [184]3 years ago
4 0

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

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If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac
MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

5 0
3 years ago
Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca
Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

k=Aexp(-Ea/RT)\\

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}

Ea_1-Ea_2 is the  lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

\frac{k_2}{k_1} =1\times 10^5

ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ

4 0
3 years ago
You know that breaking up the tablet increases the surface area of the seltzer. So what could you conclude about the relationshi
oksano4ka [1.4K]

Answer:

Grinding or breaking an Alka-Seltzer tablet increases the number of particles and increases the surface area. Material which was within the tablet is exposed, allowing for more collisions between reactant particles and resulting in an increased rate of reaction.

4 0
2 years ago
true or false: the use of fossil fuels for energy production beneficial to the environment and not cause environmental problems
MissTica
Answer:


False
Step by step
8 0
3 years ago
Heat is added to a 200.-gram sample of H2O(s) to melt the sample at 0°C. Then the resulting H2O (image) is heated to a final tem
pochemuha
You are given 200 grams of H2O(s) at an initial temperature of 0°C. you are also given the final temperature of water after heating at 65°C. You are required to get the total amount of heat to melt the sample. The specific heat capacity, cp, of water is 4.186 J/g-°C. Let us say that T1 = 0°C and T2 = 65°C. The equation for heat, Q, is  

Q = m(cp)(T2-T1)
Q = 200g(4.186 J/g-°C )(65°C - 0°C)
<u>Q = 54,418J</u>
3 0
3 years ago
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