Number 1 and 2 plz I really need it lol

n Part B, suppose the tablet was mostly dissolved when some of your solution splashed out of the beaker as CO2 continued to evol

Mandarinka [93]

Answer:

The perceived mass of CO2 would not be affected in large quantities because the splash constitutes small particles of water with sodium bicarbonate that is still reacting. The final calculated mass of sodium bicarbonate in the tablet would be artificially low.

Explanation:

Effervescence is a chemical process that involves the reaction of an acid with a carbonate or sodium bicarbonate, releasing carbon dioxide through a liquid. An example is seen in carbonated beverages, in these the gas that escapes from the liquid is carbon dioxide. The bubbles that are seen are produced by the effervescence of the dissolved gas, which by itself is not visible in its dissolved form.

8 0

3 years ago

A first order reaction has rate constants of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0ºC and 20ºC, respectively. What is the value

Airida [17]

Answer:

D. 18,800 J/mol

Explanation:

We need to use the Arrhenius equation to solve for this problem:

$k=Ae^{\frac{-E_a}{RT}$ , where k is the rate constant, A is the frequency factor, $E_a$ is the activation energy, R is the gas constant, and T is the temperature in Kelvins.

We want to find the value of $E_a$ , so let's plug some of the information we have into the equation. The gas constant we can use here is 8.31 J/mol-K.

At 0°C, which is 0 + 273 = 273 Kelvins, the rate constant k is $4.6*10^{-2}$ . So:

$k=Ae^{\frac{-E_a}{RT}$

$4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}$

At 20°C, which is 20 + 273 = 293 Kelvins, the rate constant k is $8.1*10^{-2}$ . So:

$k=Ae^{\frac{-E_a}{RT}$

$8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}$

We now have two equations and two variables to solve for. We just want to find Ea, so let's write the first equation for A in terms of Ea:

$4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}$

$A=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }$

Plug this in for A in the second equation:

$8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}$

$8.1*10^{-2}=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }e^{\frac{-E_a}{8.31*293}$

After some troublesome manipulation, the answer should come down to be approximately:

Ea = 18,800 J/mol

The answer is thus D.

5 0

3 years ago

Which atom is most likely to accept electrons to form an ionic bond?

Alinara [238K]

Halogens (atoms with 7 valence electrons) and Hydrogen

or generally, atoms with their shells almost full

5 0

4 years ago

PLS I NEED HELP ASAP! BRAINLIEST TO FULL ANSWER!!

Serggg [28]

Answer:

CARBON HYDROGEN NYTROGEN OXYGEN

Explanation:

POGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOGPOG

7 0

3 years ago

Read 2 more answers

What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

Answer: The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$