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Nimfa-mama [501]
3 years ago
7

Please help me................I beg you.........

Chemistry
1 answer:
Finger [1]3 years ago
5 0

Answer:

the problem is that Gary is not the smartest snail.the hypothesis is that he thinks he can get smarter by eating super snail food everyday.

hope this helps

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In this prelab exercise, you will calculate how to prepare solutions that you will make in lab. Below is an example of how to pr
Anastasy [175]

Explanation:

Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution (L)}}

Moles of compound =

\frac{\text{mass of compound}}{\text{Molar mass of compound}}

We have ;

Volume of solution = 600 mL = 0.600 L ( 1 mL = 0.001 L)

Moles of NaOH = n

Molarity of the solution = 3 M

3M=\frac{n}{0.600 L}

n = 3 M × 0.600 L = 1.800 mol

Mass of 1.800 mole sof NaOH :

1.800 mol × 40 g/mol = 72.0 g

Preparation:

Weight 72.0 grams of sodium hydroxide and add it to the 500 mL of volumetric flask along with some water. Dissolve the all the solute by adding small proportion of water. After the solution becomes clear make the water upto the mark of 500 ml.

Transfer the solution to a bigger beaker  and 100 mL of water more to it.

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3 years ago
How do I balance this chemical reaction ?
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Ignore my writing answer is in pictute

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3 years ago
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39. Which of the following substances are insoluble in water? Select all that apply.
Jet001 [13]
A, B, and C are insoluble in water
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If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​
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Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

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3 years ago
Describe how you might determine the m/z and relative abundance of the ions contributing to the peak at 21.876 min
aliya0001 [1]

The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.

<h3>What is relative abundance?</h3>
  • The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
  • When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
  • Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.

To learn more about relative abundance with the given link

brainly.com/question/1594226

#SPJ4

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