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2 years ago

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The speed of light in a vacuum is 2.998 × 10 8 8 m/s. How long does it take for light to circumnavigate the Earth, which has a c

ircumference of around 24,900 miles? Do not enter “seconds” as part of your answer.

1 answer:

svp [43]2 years ago

8 0

Answer:- 0.134 seconds

Solution:- The speed is given as $2.988*10^8\frac{m}{s}$ and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.

We need to do unit conversion from miles to meter as the speed is given in meter per second.

1 mile = 1609.34 meter

So, $24900mile(\frac{1609.34meter}{mile})$

= 40072566 meters

Know that, $speed=\frac{distance}{time}$

It's rearranged to time as, $time=\frac{distance}{speed}$

Let's plug in the values in it:

$time=\frac{40072566meter}{2.988*10^8\frac{meter}{second}}$

= 0.134 seconds

So, the light would take 0.134 seconds to travel the mentioned speed. The answer without the unit is 0.134.

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A graduated cylinder holds 100 mL of water. A lead weight is dropped into the cylinder bringing the new volume up to 450 mL. If

dolphi86 [110]

11.43g/mL

Explanation:

Given parameters:

Volume of water in the graduated cylinder = 100mL

Volume of water + lead weight = 450mL

Mass of lead weight = 4000g

Unknown:

Density of the lead weight = ?

Solution:

Density is the mass per unit volume of a body.

Density = $\frac{mass}{volume}$

Volume of the lead weight = volume of water displaced

Volume of lead weight = 450 - 100 = 350mL

Density = $\frac{4000}{350}$ = 11.43g/mL

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2 years ago

1. Which class of compounds contains at least one element from Group 17 of the Periodic Table ? A) aldehyde B ) amine C) ester D

blondinia [14]

Halide

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2 years ago

What is net ionic equation borax hydrolysis?

deff fn [24]

The net ionic equation of borax hydrolysis would be:

<span>Na2B4O7 + 7H2O-----------------> 2 NaOH + 4 H3BO3
</span>
Wherein 2 moles of sodium hydroxide and 4 moles of boric acid are produced by the hydrolysis.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

2 years ago

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Rama09 [41]

<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

= 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

= 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>

We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

= 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0

2 years ago

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

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