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Stolb23 [73]
3 years ago
10

When rubbed against a silk cloth, a glass rod loses electrons. How does this affect the charges of the glass rod and the silk cl

oth?
A.
The glass rod becomes positively charged, and the silk cloth becomes negatively charged.
B.
The glass rod becomes negatively charged, and the silk cloth becomes positively charged.
C.
The glass rod and the silk cloth both become positively charged.
D.
The glass rod and the silk cloth both become negatively charged.
Chemistry
1 answer:
AURORKA [14]3 years ago
7 0

Answer:

Explanation:

The glass rod losses electrons because the silk cloth has a positive charge so it attracts the negative charge of the glass rod.

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What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen to make ammonia
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If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.

We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
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3 years ago
Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after
Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is 2.83\times 10^{22}

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the air = 37^oC=[37+273]K=310K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of air = ?

Putting values in above equation, we get:

1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol

Hence, the number of moles of air present in the RV is 0.047 moles

  • <u>For b:</u>

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.047 moles of air will contain (0.047\times 6.022\times 10^{23})=2.83\times 10^{22} number of gas molecules.

Hence, the number of molecules of gas is 2.83\times 10^{22}

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