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2 years ago

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A graduated cylinder (approximate as a regular cylinder) has a radius of 1.045 cm and a height of 30.48 cm. What is the volume o

f the cylinder in cm3 (use 3.141 for TT)? Round to the nearest tenths.

1 answer:

Stels [109]2 years ago

6 0

Answer: 104.5 cm^3

Explanation:

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

Observation And Assessment

Sloan [31]

I'm confused here??????

5 0

2 years ago

Find a part of the article that describes signals that are sent within Diego’s body. Where does the signal come from, and how do

ycow [4]

Answer:

The sensory receptors send signals to Diego's brain cells. These signals are messages that help Diego figure out what to do next. As Diego thinks, more signals move from one brain cell to another.

6 0

1 year ago

What is the empirical formula of a compound which contains 84.4% c and 15.6% h by mass?

podryga [215]

Answer: The empirical formula for the given compound is $CH_2$

Explanation : Given,

Percentage of C = 84.4 %

Percentage of H = 15.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 84.4 g

Mass of H = 15.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{84.4g}{12g/mole}=7.03moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{15.6g}{1g/mole}=15.6moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.

For Carbon = $\frac{7.03}{7.03}=1$

For Hydrogen = $\frac{15.6}{7.03}=2.22\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_1H_2=CH_2$

7 0

2 years ago

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h

frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0

2 years ago