All categories

3 years ago

For many purposes we can treat nitrogen as an ideal gas at temperatures above its boiling point of - 196.°C.

Chemistry

2 answers:

Slav-nsk [51]3 years ago

4 0

Answer:

Explanation:

P1V1=n1RT1

P2V2=n2RT2

n1=n2 R=R

Divide equals and get

P1V1T2=P2V2T1

V2 =V1 (P1/P2)(T2/T1)

V2 will reduce by 10% due to pressure

V2 will increase by 184K/175K or 5%

therefor overall V will decrease

about 5%

Pavel [41]3 years ago

3 0

Answer: the volume of the sample decreased

Explanation:

T1 = -98°C = - 98 + 273 = 175K

T2 = -89°C = -89 +273 = 184K

P1 = P

P2 = 110%P = 1.1P

V1 = V

V2 =?

P1V1/T1 = P2V2/T2

PxV/175 = 1.1PxV2/184

175x1.1PxV2 = PVx 184

V2 = (PVx 184) /(175x1.1P)

V2 = 0.96V = 96%V

Therefore, the final volume is 96% of the initial volume. This means that the final volume decreased by 96%

You might be interested in

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers

A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and

Lunna [17]

The pH of the unknown solution is 3.07.

Explanation:

1.Find the cell potential as a function of pH

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0 lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

2. Find the Q value

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse) × 298.15

K × 2pH /2×96 485 C⋅mol

( inverse)

E cell= 0.0592 V × pH

3. Finding the pH value

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0

3 years ago

How do metalloids affect modern technology?

Lady bird [3.3K]

examples of metalloids :
Boron,Silicon ,Germanium are some metalloids ...
what they do:
they are used to form the semiconductors and these semiconductors are used in modern computer technology like in circuits ,chips and computer based gadgets... :)

7 0

3 years ago

Calculate the number of moles equivalent to 12.7 gram of iodine molecule

Alexus [3.1K]

$\LARGE{ \boxed{ \purple{ \rm{Answer}}}}$