Which of the following represents the atomic number of an element?

Methyl phenolphthalein and litmus

Butoxors [25]

Answer:

Both are the indicators which helps to identity the acid and base

4 0

3 years ago

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For a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k, calculate the inner diameter if you are designing for

bonufazy [111]

The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m

<h3>What is Stack Height ?</h3>

Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.

If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.

Given is a steel stack that exhausts 1,200 cu.m/min of gases

P= 1 atm and

T= 400 K

maximum expected wind speed at stack height of 12 m/s

The formula for the diameter of chimney

$\rm d=\sqrt{\dfrac{4Q}{\pi v} }$

Q =1200 cu.m/min

= 1200 * 0.0166 = 19.92 cu.m/sec

Velocity = 12m/s

$\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\$

d= 1.45 m

Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.

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8 0

2 years ago

Volume of 14.00g of nitrogen at 5.64atm and 315k

VikaD [51]

The volume of a 14.00g of nitrogen at 5.64atm and 315K is 4.59L.

<h3>How to calculate volume?</h3>

The volume of an ideal gas can be calculated using the following ideal gas equation formula;

PV = nRT

Where;

P = pressure (atm)
V = volume (L)
n = number of moles
R = gas law constant
T = temperature

An ideal gas is a hypothetical gas, whose molecules exhibit no interaction, and undergo elastic collision with each other and with the walls of the container.

The number of moles in 14g of nitrogen can be calculated as follows:

moles = 14g ÷ 14g/mol = 1mol

5.64 × V = 1 × 0.0821 × 315

5.64V = 25.86

V = 25.86 ÷ 5.64

V = 4.59L

Therefore, 4.59L is the volume of the gas

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4 0

1 year ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

7 0

3 years ago

During science class, while studying mixtures, you mix together table salt and sand. Your teacher challenges you to separate the

olga2289 [7]

The property to be used is sieving. you get a bowl and a sieve when you pour the sand and salt,the salt being the small particle sieve to the bowl while the sand remain on the sieve as residue

7 0

3 years ago