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3 years ago

5

you need to measure the diameter of a soda can. which measuring tool would produce accurate result that you would expect to be t

he same as those found by others

1 answer:

snow_lady [41]3 years ago

5 0

A ruler maybe? im not really sure srry

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statuscvo [17]

Answer:

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3 years ago

This chart shows descriptions of the atoms for four different substances.

Whitepunk [10]

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3 years ago

A sample of 53.0 of carbon dioxide was obtained by heating 1.31 g of calcium carbont. What is the percent yield for this reactio

MaRussiya [10]

Answer:

92.04%

Explanation:

Given:

Mass of CO₂ obtained = 53.0 grams

Mass of calcium carbonate heated = 1.31 grams

Now,

the molar mass of the calcium carbonate = 100.08 grams

The number of moles heated in the problem = Mass / Molar mass

= (1.31 grams) / (100.08 grams/moles)

= 0.013088 moles

now,

1 mol of calcium carbonate yields 1 mol of CO₂

thus,

0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂

now,

Theoretical mass of 0.013088 moles of CO₂ will be

= Number of moles × Molar mass of CO₂

= 0.013088 × 44 = 0.5758 grams

Thus, the percent yield for this reaction = $\frac{\textup{Actual yield}}{\textup{Theoretical yield}}\times100$

or

the percent yield for this reaction = $\frac{0.53}{0.5758}\times100$

or

the percent yield for this reaction = 92.04%

6 0

3 years ago

31) give an example of one renewable and one non renewable resource. be sure to state which is renewable and which is nonrenewab

Yakvenalex [24]

31) A renewable resource is qualified as things like, Solar, Wind, Hydro-electric, and thermal.
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4 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

3 years ago