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olasank [31]
3 years ago
7

Use the helium-4 isotope to define atomic number and mass number. why does knowledge of atomic number enable us to deduce the nu

mber of electrons present in an atom
Chemistry
1 answer:
Nuetrik [128]3 years ago
8 0
Helium - 4 isotope has 2 protons and 2 neutrons.

Its atomic number is 2 (the number of protons). Atomic number is the number of protons in the atom of an element and it identifies the element. All the isotopes of an element have the same atomic number (same number of protons). So, all the isotopes of hellium have 2 protons.

The number 4 that identifies the helium - 4 isotope is the mass number, i.e. the number of protons plus the number of neutron. So, helium-4 isotope has 2 protons and 2 neutrons. While, other isotopes of helium have 2 protons but different number of neutrons.

The atomic number enable us to deduce the number of electrons in an atom because the atoms are neutral, which implies that the positive and negative charge are equal. Given that the magnitude of the charge of electons and protons are equal but the sign is opposed (protons are positve and electrons are negative) the number of electrons in the atom equals the number of protonts. That is why the atomic number is also the number of electrons of the atom.
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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s
Slav-nsk [51]
Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
6 0
3 years ago
I need help TnT
Slav-nsk [51]

Explanation:

I don't think human beings will ever extinct

cause they are smart and adaptive

7 0
2 years ago
Read 2 more answers
Describe the four types of fronts and the weather associated with each.
FrozenT [24]

The Four type is

Cold front

Warm front

Stationary front

Occluded front

3 0
3 years ago
A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?
Oksana_A [137]
Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

\frac{V1}{T1} = \frac{V2}{T2} \\  \\ T2= \frac{V2*T1}{V1}=  \frac{235 mL * 105  ^{\circ}C }{455 mL}=54,23 ^{\circ}C

So, the final temperature is 54,23 °C

Have a nice day!
5 0
3 years ago
Read 2 more answers
Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined
Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

3 0
3 years ago
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