All categories

2 years ago

Two sealed flasks each have a volume of 2.0 liters. Flask A contains N, O gas and Flask B contains H, gas both at 1

Chemistry

1 answer:

barxatty [35]2 years ago

4 0

Answer:

Each gas have same number of molecules.

Explanation:

According to Avogadro law,

Equal volume of all the gases at same temperature and pressure have equal number of molecules.

Mathematical expression:

V ∝ n

V = Kn

V/n = K

k = constant

V = volume of gas

n = number of moles of gas

when volume change is changed from v1 to v2 and number of moles from n1 to n2 this law can be written as,

V1 / n1 = V2 /n2

This state that by increasing the number of moles of gas volume also goes to increase.

You might be interested in

A plastic bin is found to hold 3.1x10^24 molecules of water.

motikmotik

Answer:

$\boxed {\boxed {\sf 5.1 \ mol \ H_2O}}$

Explanation:

To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.

We can use it in a ratio.

$\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}$

Multiply by the given number of molecules.

$3.1*10^{24} \ molecules \ H_2O*\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}$

Flip the ratio so the molecules of water cancel out.

$3.1*10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O}{6.02*10^{23} \ molecules \ H_2O}$

$3.1*10^{24} *\frac {1 \ mol \ H_2O}{6.02*10^{23} }$

$\frac {3.1*10^{24} \ mol \ H_2O}{6.02*10^{23} }$

Divide.

$5.14950166113 \ mol \ H_2O$

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

$5.1 \ mol \ H_2O$

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.

5 0

2 years ago

Please help!!! (im really bad at chemistry)

Lubov Fominskaja [6]

Answer:

1000ml/1L

Explanation:

8 0

1 year ago

2. Which of the following elements does not lose an electron easily? (a) Na (6) F (c) Mg (d) Al

vaieri [72.5K]

Answer:

Answer: (b) F

Explanation:

Sodium has 1, magnesium has 2 and Aluminium has 3 electrons in its outermost shell whereas Fluorine has 7 electrons in its outermost shell hence Fluorine does not lose electrons easily.

The electronic configuration of fluorine is 2,7.

Fluorine is the ninth element with a total of 9 electrons.

The first two electrons will go in the 1s orbital.

The next 2 electrons for F go in the 2s orbital.

The remaining five electrons will go in the 2p orbital. Therefore the F electron configuration will be 1s22s22p5.

3 0

1 year ago

Deposition can occur where wind or running water slows down? true or false

Shalnov [3]

Answer:

The answer is True.

Explanation:

6 0

2 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

2 years ago