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USPshnik [31]
3 years ago
10

How many atoms of phosphorus are in a 0.859 g sample? Answer in units of atoms.

Chemistry
1 answer:
bekas [8.4K]3 years ago
3 0
Moles = Mass(g) / ar
Moles of phosphorus = 0.859 / 31
= 0.0277096774

1 mole = 6.02 × 10^23

Atoms of phosphorus = 0.0277096774 × 6.02×10^23
=ans.

sorry i cant do the calculation. I dont have a calculator with me.
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Which of these gases will have the greatest density at the same specified temperature and pressure?
ch4aika [34]

Answer:

Neon

The highest density among the inert gases is of Neon (Ne). This a factual data. Thus the highest density among given options is of Ne, as all the options are of inert gases.

Explanation:

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5 0
3 years ago
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati
babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l),  ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. <u>Therefore, the given reaction is exothermic.</u>

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

<u>Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.</u>

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. <u>Therefore, the given combustion reaction is an oxidation-reduction reaction.</u>

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

<u>Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.</u>

3 0
3 years ago
Carbon-13 is an isotope. It has 6 protons and _____. 13 neutrons 7 neutrons 7 electrons 13 electrons
vovangra [49]

Answer:

7 neutrons

Explanation:

7 0
3 years ago
Read 2 more answers
7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s
Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both NH_{3} and CO_{2} are gaseous. Hence equilibrium constant depends upon partial pressures of NH_{3} and CO_{2}.

Initially no NH_{3} and CO_{2} were present.

Hence mole fraction of NH_{3} and CO_{2} at equilibrium can be calculated from coefficient of NH_{3} and CO_{2} in balanced equation.

Mole fraction of NH_{3} = (number of moles of NH_{3})/(total number of moles of NH_{3} and CO_{2}) = \frac{2moles}{(2+1)moles}=\frac{2}{3}

Mole fraction of CO_{2} = (number of moles of CO_{2})/(total number of moles of NH_{3} and CO_{2}) = \frac{1moles}{(2+1)moles}=\frac{1}{3}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
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