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Ainat [17]
3 years ago
10

How many moles are in 98.3 grams of NaF?

Chemistry
2 answers:
boyakko [2]3 years ago
5 0

Answer:

1 mole is equal to 6.022x10²³

Anton [14]3 years ago
3 0

Answer:

2.979 mol

Explanation:

mass of NaF=14+19

                    =33g

33g of NaF is the mass of 1 mol

98.3g of NaF is the mass of (98.3/33)*1=2.979 mol

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Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha
DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

CaO(s)+CO_2\rightarrow CaCO_3(s)     \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)    \Delta H_1=-812.8kJ  

(2) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)     \Delta H_2=-1269kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0
3 years ago
In this experiment, you were asked to calculate your percent yield of alum assuming that you started with pure aluminum. Actuall
neonofarm [45]

Answer:

Percent yield of alum is 38.6%

Explanation:

Find attached calculations for detailed explanation.

3 0
3 years ago
What is the epicenter?
hoa [83]

Answer:

basically the middle of an earthquake

5 0
3 years ago
In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
3 years ago
Carbon has an atomic number of 6. How many electrons are in a carbon atom’s second energy level?
matrenka [14]
Carbon is an atom so electrons=protons 
<span>First shell has 2 electrons and the next 8 so with 6 electrons the first shell fills and 4 are left to go into the 2th shell.</span>
8 0
3 years ago
Read 2 more answers
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