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3 years ago

Make the following conversion. 0.075 m = _____ cm 7.5 75 0.0075 0.00075

1 answer:

8 0

In the given question we need to convert 0.075 meter into centimeter. For this purpose we need to know the relation between meter and centimeter. Then only this problem can be solved and the correct answer reached.
Now we know that
1 meter = 100 centimeter
Then
0.075 meter = [(100/1) * 0.075] centimeter
We know that any number when divided by 1 gives the same number.
So
0.075 meter = (100 * 0.075) centimeter
= (100 * 75/1000) centimeter
= (7500/1000) centimeter
= (75/10) centimeter
= 7.5 centimeter.
So from all the given options the correct answer to the question is 7.5 centimeter.

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George has a mass of 65kg. What would George weigh on the Moon? (The gravitational field strength on the Moon is 1.6N/kg.)

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Answer:

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Explanation:

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How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe

polet [3.4K]

First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

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3 years ago

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What’s the question

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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.

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