Repeated trials and replication are the same thing. - True

How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?

statuscvo [17]

Answer:

A) $6.5\times 10^{23}\ \text{molecules}$

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

The reaction of electrolysis would be

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Number of moles of $H_2O$

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}$

From the reaction it can be seen that 2 moles of $H_2O$ gives 1 mole of $O_2$

So, number of moles of $O_2$ produced is

$\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}$

Number of molecules

$1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}$

So, $6.5\times 10^{23}\ \text{molecules}$ of oxygen is produced.

8 0

2 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

The equation below shows the products formed when a solution of silver nitrate (AgNO3) reacts with a solution of sodium chloride

d1i1m1o1n [39]

I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant

3 0

3 years ago

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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g

ArbitrLikvidat [17]

KClO3(s)+Δ→KCl(s)+32O2(g)

Approx. 3L of dioxygen gas will be evolved.

Explanation:

We assume that the reaction as written proceeds quantitatively.

Moles of KClO3(s) = 10.0⋅g122.55⋅g⋅mol−1 = 0.0816⋅mol

And thus 32×0.0816⋅mol dioxygen are produced, i.e. 0.122⋅mol.

At STP, an Ideal Gas occupies a volume of 22.4⋅L⋅mol−1.

And thus, volume of gas produced = 22.4⋅L⋅mol−1×0.0816⋅mol≅3L

Note that this reaction would not work well without catalysis, typically MnO2.

8 0

3 years ago

Read 2 more answers

Which sub-layer thins out into space, where there is no air

KengaRu [80]

The "exosphere" is the most distant and tenuous "layer" of our atmosphere.

3 0

2 years ago

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Repeated trials and replication are the same thing. - True - False