Explanation:
Examples of chemical changes include, rusting, fire, and overcooking. Nuclear changes occur when the nuclei of atoms are rearranged to form new atoms. Examples of nuclear changes include atomic fission, nuclear fusion, and the energy of sun and stars.
Answer:
Explanation:
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In this case, according to the given information, it turns out possible to set up the initial version of the half-reaction as follows:
Now, we assign the oxidation numbers to both nitrogen atoms:
It means three electrons are carried, two water molecules are needed on the left side in order to balance the oxygen atoms and consequently four hydronium ions to balance hydrogen atoms in acidic media:
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The balanced equation is Mg + 2AgNO₃ ⟶ Mg(NO₃)₂ + 2Ag
Step 1. Write the <em>unbalanced equation
</em>
Mg + AgNO₃ ⟶ Mg(NO₃)₂ + Ag
Step 2. Start with the<em> most complicated-looking formula</em> [Mg(NO₃)₂] and balance its atoms.
Mg: Already balanced —1 atom each side.
N: We need 2 N on the left. Put a 2 in front of AgNO₃.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + Ag
O: Already balanced —6 atom6 each side.
Step 3: Balance <em>Ag</em>
We have 2Ag on the left. We need 2Ag on the right.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + 2Ag
Answer: A woman kicks a soccer ball and scores a goal.
Explanation:
Potential energy: it is the energy possessed by the body due to its position.
Kinetic energy: it is the energy possessed by the the body due to its motion.
When a woman kick's a soccer ball she transferred her potential energy to soccer ball.Due to this action soccer ball comes into motion which means potential energy imparted to the ball starts getting converted into kinetic energy.
Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
