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2 years ago

Which of the following would be a warning sign that information about a weight loss medication is not reliable science?

Chemistry

1 answer:

Helga [31]2 years ago

5 0

Answer:

Explanation:

I think it is right not sure but hope it helped

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When chalk is added to water and stirred what is formed

lianna [129]

The chalk will dissolve

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2 years ago

Read 2 more answers

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

stealth61 [152]

Answer: The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

Explanation:

We are given a function that calculates the amount of sample remaining after 't' years, which is:

$A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}$

For t = 20 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}$

$A_t(t)=288.522g$

Hence, the amount of sample left after 20 years is 288.522 g

For t = 50 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}$

$A_t(t)=144.26g$

Hence, the amount of sample left after 50 years is 144.26 g

6 0

3 years ago

Solve the problem for the moles of oxygen. mol O2 DONE

Elza [17]

Answer: The answer is 0.245

Explanation: That is what it said the answer was

8 0

2 years ago

What is the gram-formula mass of Ca3(PO4)2

Sindrei [870]

310.17 grams so there

4 0

3 years ago

N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f

kow [346]

Answer : The final concentration of $N_2O_5$ is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $5.89\times 10^{-3}\text{ min}^{-1}$

t = time passed by the sample = 3.5 min

a = initial concentration of the reactant = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

$3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}$

$a-x=2.9M$

Thus, the final concentration of $N_2O_5$ is, 2.9 M

3 0

2 years ago