Answer:
The Atomic Number of the atom of an element whose model is given is "8" that is option no. 'C' in the question.
Explanation:
An Atom comprises of 3 basic structures that are Protons, Neutrons and Electrons. The central part is the Nucleus which contains protons and neutrons having positive charge and no charge respectively. The electrons are revolving around the nucleus in electronic shells having the negative charge.
<u><em>ATOMIC NUMBER: </em></u>
Atomic number is the number of protons present inside the nucleus of an atom and it determines the place of that particular atom in the <u>Periodic Table.</u>
In the model, given in the question, the nucleus contains 2 types of balls dark gray colored and light gray colored. The key at the bottom shows the dark gray colored ball as having a positive charge and thus it represents the atomic number for the given atom of element which is <u><em>EIGHT (8).</em></u>
So, the atomic number for the given atom is 8 which is element OXYGEN.
Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
Hello!
To solve this problem we are going to use the
Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is
4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is
5,75Have a nice day!
Answer: The value of 
for the reaction is, -2512.4 kJ
Explanation:
The chemical equation for the combustion of acetylene follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(4\times (-393.5))+(2\times (-241.8))]-[(2\times (227.4)+(5\times (0))]\\\\\Delta H^o_{rxn}=-2512.4kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%284%5Ctimes%20%28-393.5%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28227.4%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-2512.4kJ)
Therefore, the value of for the reaction is, -2512.4 kJ
