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2 years ago

What are 2 examples of what happens when scientist do not have a standard system of measurement of the scientific community?

Chemistry

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

Explanation:

The standard system of measurement is what is popularly referred to as SI unit (or international system of units). For example, the SI unit for length is basically in meters (m), which is convertible to millimeter (mm), centimeter (cm) or even kilometer (km).

If there is no standard system of measurement,

1) Scientists will not be able to compare data as every scientist will use a unit that suits him/herself which might make data incomparable as some units might be difficult to convert to other units or there conversion methods might even be debatable.

2) There might be a general lack of understanding of the research work done by a scientist/scientists in the scientific community. This is because the standard system of measurement affords scientists to communicate properly with the same or convertible units and based on 1. above, if scientists use any unit that isn't agreed upon, different/unknown/debatable units might make understanding of research works incomprehensible.

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3 years ago

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

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3 years ago

Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r

MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

O = 53.29/16 = 3.33

H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

O = 3.33/3.33= 1

H = 6.71/3.33 = 2

The empirical formula of the compound ribose is CH2O

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3 years ago

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Answer:

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