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3 years ago

Help! Will mark brainliest! Explain why a mechanical wave does not exist in a vacuum.

1 answer:

7 0

<span> Light waves are a wave in the electromagnetic field, this field may exist in a vacuum. Sound needs either solid, liquid, or gas to propagate and therefore it is called a mechanical wave. Sound is a form of energy which is transferred by vibrating air molecules if motions of propagation is air.</span>

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How many oxygen atoms are in 1.40 g of quartz?

PtichkaEL [24]

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4 0

3 years ago

Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h

Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

$k=\frac{0.693}{t_1_/_2}$

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

$k=\frac{0.693}{1.5 hrs}$

The expression for the rate constant is :

$k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}$

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

$\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6$

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0

3 years ago

Describe professions for chemists

zepelin [54]

Answer:

A chemist searches for new knowledge about chemicals and use it to improve the way we live. He or she may develop products such as synthetic fibers, drugs and cosmetics. Chemists create processes, including all refining and petrochemical processing, that reduce energy use and pollution.

8 0

3 years ago

There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen

alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

8 0

3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

3 years ago

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