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lutik1710 [3]
3 years ago
9

Do you think that it is necessary and/or important to take a multivitamin everyday to maintain optimal health? What might help y

ou answer this question? What types of things might influence your decision to take a multivitamin every day?
What's ur opinion with explanation???????!!!!!!!!!!
Chemistry
1 answer:
kolbaska11 [484]3 years ago
8 0
I do not think it is necessary to take a multivitamin everyday. To maintain optimal health, people should eat a balanced diet with low processed food and stay active. Looking into your daily health habits can help determine if you want to take multivitamins. Factors such as level of activity, diet, time outside, and physical condition can influence this decision.
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Tim found a battery powered hand-held fan. He says that the fan must have an electric motor to power the blades of the fan. Ahis
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Answer:

fan is a powered machine used to create a flow of air. A fan consists of a rotating arrangement of vanes or blades, ... Most fans are powered by electric motors

5 0
2 years ago
A container holds a pure sample of a radioactive substance with a half-life of 2 hours. Which of the following statements are tr
rodikova [14]

C, because a half-life is the amount of time it takes for half the atoms to decay.

5 0
4 years ago
A fast moving object goes a long distance in a _______ period of time
harina [27]
I believe the answer you are looking for is : SHORT
4 0
4 years ago
Read 2 more answers
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sampl
Rzqust [24]

Answer:

0.193 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 0.250 L

Initial temperature (T1) = 27°C

Initial pressure (P1) = 0.850 atm

Final volume (V2) =?

Final temperature (T2) = 0°C

Final pressure (P2) = 1.00 atm

Step 2:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 27°C = 27°C + 273 = 300K

Final temperature (T2) = 0°C = 0°C + 273 = 273K

Step 3:

Determination of the new volume of the sample of ammonia gas.

The new volume can be obtain by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

0.850x0.250/300 = 1xV2/273

Cross multiply to express in linear form

300 x V2 = 0.850x0.250x273

Divide both side by 300

V2 = (0.850x0.250x273) /300

V2 = 0.193 L

Therefore, the volume of the sample at 0 °C and 1.00 atm is 0.193 L

6 0
3 years ago
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T
motikmotik

Answer :

(a) The rate law for the reaction is:

\text{Rate}=k[OCl^-]^1[I^-]^1

(b) The value of rate constant is, 60.4M^{-1}s^{-1}

(c) rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

OCl^-+I^-\rightarrow OI^-+Cl^-

Rate law expression for the reaction:

\text{Rate}=k[OCl^-]^a[I^-]^b

where,

a = order with respect to OCl^-

b = order with respect to I^-

Expression for rate law for first observation:

1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b ....(1)

Expression for rate law for second observation:

2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b ....(2)

Expression for rate law for third observation:

2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b ....(3)

Dividing 1 from 2, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1

Dividing 1 from 3, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1

Thus, the rate law becomes:

\text{Rate}=k[OCl^-]^a[I^-]^b

a  = 1 and b = 1

\text{Rate}=k[OCl^-]^1[I^-]^1

Now, calculating the value of 'k' (rate constant) by using any expression.

1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})

k=60.4M^{-1}s^{-1}

Now we have to calculate the rate for a reaction when concentration of OCl^-  and I^-  is 1.8\times 10^{-3}M and 6.0\times 10^{-4}M respectively.

\text{Rate}=k[OCl^-][I^-]

\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)

\text{Rate}=6.52\times 10^{-5}Ms^{-1}

Therefore, the rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

8 0
3 years ago
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