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2 years ago

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What is the empirical formula for butane, C4H10

1 answer:

pochemuha2 years ago

4 0

C2H5 i think by ÷2 on both

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Identify each of the following sets of quantum numbers as allowed or not allowed in the hydrogen atom.

Westkost [7]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

$1s^1$

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.

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2 years ago

PLS I NEED HELP with steps WILL GIVE BRAINLIEST

RideAnS [48]

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3 years ago

Ca (ClO) 2<br> Spell out the full name of the compound.

antoniya [11.8K]

Calcium hypochlorite

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3 years ago

Calculate the amount of heat required to melt 2108 g of water. The enthalpy of fusion of water is ΔHfus=6.010 kJ/mol.

vesna_86 [32]

since the unit for the heat of fusion is kJ/mol, you're going to have to convert the grams into moles in order to cancel out the unit. After that, you can solve like normal.

6 0

2 years ago

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

<u>Part A:</u>

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

<u>Part B:</u>

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0

2 years ago