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4 years ago

How many electrons in an atom can share the quantum numbers n = 1, m subscript l = 0? A. 1 B.2 C.3 D.6

Chemistry

1 answer:

Kisachek [45]4 years ago

6 0

I believe it would be c: 3 because you can have the numbers -1,0,1.

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How is the outer core a liquid?

Nata [24]

The outer core is not under enough pressure to be a solid so it is liquid

3 0

4 years ago

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

Vladimir [108]

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

Second, we gather the information what we are going to use in our calculations.

Volumen of Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 = 0.541M

Molecular Weight Al2O3 = 101.96 g/mol

Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

$\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 = 0.022moles Al(NO3)3$

then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced

$0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3$

finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.

$0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3$

4 0

4 years ago

What is the density of platinum

Alex Ar [27]

Answer:

21.45 g/cm3

Explanation:

6 0

3 years ago

Read 2 more answers

The mineral vanadinite has the formula pb5(vo4)3cl. what mass percent of chlorine does it contain?

lubasha [3.4K]

The molar mass of vanadinite is 1416.27 g/mol. Looking at the molecular formula we can see that there is 1 mole of Cl for every mole of vanadinite. Therefore:

mass percent = [(1 mole Cl * 35.45 g/mol) / (1 mole Pb5(VO4)3Cl * 1416.27 g/mol)] * 100%

<span>mass percent = 2.50%</span>

5 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago