Answer:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Explanation:
Chemical equation:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Balance chemical equation:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Ionic equation
Ag⁺(aq)+ NO₃⁻(aq) + Na⁺(aq)+ Cl⁻(aq) → AgCl(s) + Na⁺(aq)+ NO₃⁻(aq)
Net ionic equation:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
The NO₃⁻((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The AgCl can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
Q = 30355.2 J
Explanation:
Given data:
Mass of ice = 120 g
Initial temperature = -5°C
Final temperature = 115°C
Energy required = ?
Solution:
Specific heat capacity of ice is = 2.108 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 -T1
ΔT = 115 - (-5°C)
ΔT = 120 °C
Q = 120 g × 2.108 j/g.°C × 120 °C
Q = 30355.2 J
The new volume of a gas at STP that occupies 16L of space is 27.43L.
<h3>HOW TO CALCULATE VOLUME?</h3>
The volume of a given gas can be calculated using the following expression:
P1V1/T1 = P2V2/T2
Where;
- P1 = initial pressure
- P2 = final pressure
- V1 = initial volume
- V2 = final volume
- T1 = initial temperature
- T2 = final temperature
1 × 16/273 = 2.6 × V2/468
16/273 = 2.6V2/468
0.0586 × 468 = 2.6V2
V2 = 27.43
Therefore, the new volume of a gas at STP that occupies 16L of space is 27.43L.
Learn more about volume at: brainly.com/question/1578538
Answer:
Viscosity of water decreases markedly with increasing temperature. When the viscosity decreases, the flow resistance decreases. So for the same driving force, that is the pressure drop per unit length, the water flow rate will be higher.
Hope this helps :)
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
