Answer: atoms do have 3 subatomic particles but the nucleus is positive and the nucleus consists of protons and neutrons. Its positive because neutrons have no charge and protons have a positive charge. There are only electrons on the shells so no neutrons or protons on the shells
Explanation:
6.349 g mass of anhydrous magnesium sulfate will remain.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Molar mass MgSO₄.7 H₂O = 246.52 g/mol
0.0527 moles
Molar mass MgSO₄ = 120.4 g/mol
Mass of anhydrous magnesium sulfate :
( 0.0527 x 120.4 ) => 6.349 g
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One possible answer could be that a chemical reaction has occurred.
Answer:
A 12 oz Coca Cola contains 39g of sugar or C6H12O6.
To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:
39g/ 180.16 g/mol = 0.216 mol sugar
then, convert 12 oz to L:
12oz / (1oz/0.02957L) = 0.35484 L
therefore the concentration of sugar in the soda is:
M = mol sugar / L sol'n
= 0.216 mol sugar / 0.35484 L
= 0.609 M
Explanation:
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.
