<span>Answer: option (4) the same magnitude and the opposite sign.
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Justification:
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</span><span>1) Electrons are negative particles thar are around the nucleus of the atom (in regions called orbitals).
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2) Protons are positive particles that are inside the nuclus of the atom.
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<span>3) The nucleus of the atom has the same number of protons as electrons are in the orbitals of the atom.
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4) The atoms are neutral (neither positive nor negative) because there are the same number of electrons and protons and their charge are of the same magnitude but different sign: (+) + (-) = 0: positive + negative = neutral.</span>
Answer:
2.5 × 10² ppm
Explanation:
Step 1: Given data
- Mass of the sample: 200. g
Step 2: Convert 0.050 g to μg
We will use the conversion factor 1 g = 10⁶ μg.
0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg
Step 3: Calculate the concentration of NaCl in ppm
The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.
5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles reacted - 0.200 mol/L x 0.0250 L = 0.005 mol
according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted
therefore number of HCl moles - 0.005 mol
volume of 30.0 mL contains 0.005 mol
therefore 1000 mL contains - 0.005 mol / 0.030 L = 0.167 M
concentration of HCl is 0.167 M
Yeah man I can help explain a little bit fits
