Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
Answer:
0.6743 M
Explanation:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:
- Molarity = moles / volume
- moles = Molarity * volume
- 0.4293 M * 39.27 mL = 16.86 mmol NaOH
<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.
Finally we <u>calculate the concentration (molarity) of acetic acid</u>:
- 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
Answer:
Because your body has built-in resistance to certain gases, no matter the size of the gas cloud.
That is why we are able to stay non-inert to these types of gases, like Carbon dioxide.
