How many electrons would be found in an isotope of silicon 29

If the outside of your flask is not dry when the first mass determination is made and it is dried for the second mass determinat

svlad2 [7]

Answer:

You'll experience a grater deviation

Explanation:

<em>You'll experience a greater deviation in your measurements, meaning your measures will have a bigger difference between them, and the greater these deviations the less accurate will be the measuring.</em> This happens mainly because you're not replicating the measurement with the exact same conditions, in one of them you'll have an extra mass from the water.

I hope you find this information useful and interesting! Good luck!

8 0

3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

6 0

3 years ago

when 10 moles of sodium reacts with excess oxygen for 60 minutes, only 3 moles of sodium remains in 1000 millimeters of H20. Wha

diamong [38]

That would be....................

8 0

3 years ago

PLEASE ANSWER ASAP (WILL GIVE BRAINLIEST)

Flauer [41]

Answer:

I think it is either A. or B.

Explanation:

(I think!)

5 0

2 years ago

Which two changes would make this reaction reactant-favored?

Vera_Pavlovna [14]

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient

in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2

As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

4 0

2 years ago