Correct answer: a. releasing CO2 that dissolves and forms acid in the oceans
The fuels used in automobiles release gases like carbon dioxide, carbon monoxide, oxides of nitrogen and sulfur. Carbon dioxide when dissolved in water forms carbonic acid. So, when the usage of cars is high, these emissions of carbon-dioxide into the atmosphere increase and this leads to the lowering of pH of the oceans as the carbon dioxide present in higher amounts in to atmosphere diffuses into the oceanic waters and form carbonic acid which makes the ocean slightly acidic.
Answer:
Possible lowest volume = 0.19 cm
Possible highest volume = 0.21 cm
Explanation:
given data
volumetric pipette uncertainty = 0.01 cm³
total volume = 0.20 cm³
solution
we will get here Possible lowest and highest volume that is express as
Possible lowest volume = total volume - uncertainty .....................1
Possible highest volume = total volume + uncertainty ....................2
put here value in both equation and we get
Possible lowest volume = 0.20 cm - 0.01 cm
Possible lowest volume = 0.19 cm
and
Possible highest volume = 0.20 cm + 0.01 cm
Possible highest volume = 0.21 cm
Answer:
Explanation:
HCl + NaOH ⟶ NaCl + H₂O
There are two energy flows in this reaction.
Heat of reaction + heat to warm water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m(HCl) = 50 g
m(NaOH) = 50 g
T₁ = 22 °C
T₂ = 28.87 °C
C = 4.18 J·°C⁻¹g⁻¹
Calculations:
m = 50 + 50 = 100 g
ΔT = 28.87 – 22 = 6.9 °C
q₂ = 100 × 4.18 × 6.9 = 2900 J
q₁ + 2900 = 0
q₁ = -2900 J
The negative sign tells us that the reaction produced heat.
The reaction produced 
.
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
While I am not the brainliest I can certainly answer.
This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.
