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3 years ago

A hardware salesman measures the mass of a box containing

Chemistry

2 answers:

sattari [20]3 years ago

8 0

Step $1$

find the mass of a single washer

Divide the total mass by the number of washers

$\frac{0.4168}{1000} =0.0004168\ Kg$

$0.0004168kg=4.168*10^{-4} kg$

Step $2$

Convert kg to mg

we know that

$1 kg=1*10^{6} mg$

$4.168*10^{-4}*10^{6} =4.168*10^{2}mg=416.8 mg$

therefore

the answer is

$416.8 mg$

marysya [2.9K]3 years ago

7 0

In this question, the box with 1000 washers has 0.4168kg mass. Then, to find the mass of a single washer you need to divide the mass with the number of washers. Then, you need to convert the unit into milligrams(1kg= 10^6mg)
It would be 0.4168 kg / 1000 * 10^6 milligram/kg= 4168mg.

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Explanation:

In a magnetic field, the radius of the charged particle is as follows.

r = $\frac{mv}{qB}$

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q = charge, B = magnetic field

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q = $\frac{mv}{rB}$

= $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

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2 years ago

Diamond cannot conduct electricity but graphite can,why

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Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

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3 years ago