Answer:
A)
B) 0.025 mol (SCN)₂
C) 2 mol (SCN)₂
D) 
E) 3.5504 mol H₂SO₄
Explanation:
2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄
A) A conversion factor could be
, as it has the units that we want to <u>convert to in the numerator</u>, and the units that we want to <u>convert from in the denominator</u>.
B) 0.05 mol NaSCN *
= 0.025 mol (SCN)₂
C) With 4 moles of NaSCN and 3 moles of MnSO₄, the<em> reactant </em>is NaSCN so we use that value to calculate the moles of product formed:
4 mol NaSCN *
= 2 mol (SCN)₂
D) 
E) 1.7752 mol MnSO₄ *
= 3.5504 mol H₂SO₄
Answer:
A
Explanation:
The only factor that would not decrease the carrying capacity of the fur seal's environment is the population of predators.
<em>The carrying capacity is defined as the maximum number of individuals of a species an environment can sustain based on the resources available. Hence, only a fluctuation in the resources available in the environment of the fur seal can affect the carrying capacity. </em>
<u>Pollution </u>and <u>destruction of the habitat of the fur seal</u> would limit the resources available to the animals and as such, limit the carrying capacity. A <u>fluctuation in the population of prey available</u> for the population of fur seals would also affect the carrying capacity.
The only factor that has no bearing on the carrying capacity of the fur seal's environment is the population of predators. Predators have the capacity to reduce the population of fur seals but not affect the carrying capacity for fur seals.
The correct option is, therefore, A.
Answer: The mass of 405 mL of ethylene glycol is 444.3 g.
Explanation:
Density is defined as the mass contained per unit volume.
Given :
Density of the ethylene glycol=
Volume of the ethylene glycol=
= 0.405 L (1L=1000ml)
Mass of the ethylene glycol = ?
Putting in the values we get:
Thus the mass of 405 mL of ethylene glycol is 444.3 g
<h3>
Answer:</h3>
134 atm
<h3>
Explanation:</h3>
- Based on the pressure law, the pressure of a gas varies directly proportionally to the absolute temperature at a constant volume.
- Therefore; we are going to use the equation;

In this case;
Initial pressure, P1 = 144 atm
Initial temperature, T1 (48°C) = 321 K
Final temperature, T2 (25°C) = 298 K
We need to find the final pressure,
Therefore;
P2 = (P1/T1)T2
= (144/321)× 298 K
= 133.68 atm
= 134 atm
Therefore, the new pressure will be 134 atm.