Which gas variable has to do with the amount of collisions of gas particles has?

I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w

trasher [3.6K]

The percentage of yield was 777.78%

Explanation:

We have the equation,

Be [s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑ Be (s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get x 100

=3.5 g/0.45 g 100

= 777.78 %

The percentage of yield was 777.78%

5 0

3 years ago

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

Answer: The molality of solution is 0.740 m.

Explanation:

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

Volume of water = 750 mL

Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago

Explain the difference between rotation and revolution.

aleksandrvk [35]

Answer:

Rotation: spins on its axis

Revolution: travels around another object

Explanation:

The Earth rotates on its axis, while it revolves around the sun.

The moon revolves around the Earth.

8 0

3 years ago

Read 2 more answers

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

3 years ago

Gabe amodeo, a nuclear physicist, needs 8080 liters of a 3030% acid solution. he currently has a 20 %20% solution and a 60 %6

MArishka [77]

i really dont knowb lfre,;'r;'frke';few frq3wklrndklqwef2

5 0

3 years ago