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3 years ago

Complete the passage.

Chemistry

2 answers:

EastWind [94]3 years ago

4 0

Answer:

Sound waves and some earthquake waves are longitudinal waves.Ocean, light, and other earthquake waves are transverse waves

Explanation:

Natalka [10]3 years ago

3 0

Answer:

Longitudinal and Transverse

Explanation:

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NaHCO3 is an example of -

bezimeni [28]

Answer:

sodium bicarbonate

Explanation:

8 0

3 years ago

How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh

zysi [14]

Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

CO + 2 H₂   → CH₃OH

Calculating Moles of CO:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO

Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g

X = 11.25 Moles of CO

Calculating Moles of H₂:
According to equation,

32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂

Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g

X =  22.5 Moles of H₂

Result:
  3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.

3 0

3 years ago

Hi kay my account got deleted this is my new one

jek_recluse [69]

Answer:

hUh

Explanation:

3 0

3 years ago

Read 2 more answers

A piston in a heat engine does 500 joules of work, and 1,400 joules of heat are added to the system. Determine the change in int

Ira Lisetskai [31]

The change is that the piston gets hotter? (Theres more heat in it)

8 0

3 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>

6 0

3 years ago