Answer:
B
Explanation:
Not sure, but that one makes the most since
Answer:
87.3 calories of heat is required.
Explanation:
Heat = mcΔT
m= mass, c = specific heat of silver, T = temperature
H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)
H = 57.8 * 0.057 * 26.5
H = 87.3069 cal.
The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.
Answer:
Nitrogen (ii) oxide
Explanation:
To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.
NOTE: The oxidation number of oxygen (O) is always – 2.
Thus the oxidation number of N in NO can be obtained as follow:
N + O = 0 (ground state)
N + (– 2) = 0
N – 2 = 0
Collect like terms
N = 0 + 2
N = +2
Thus, the oxidation number of Nitrogen (N) in NO is +2.
Therefore, the IUPAC name for NO is Nitrogen (ii) oxide
Answer:
A more dense plate going underneath a less dense plate.
Answer:
D. 1.48atm
Explanation:
Van der waals equation is given as:
(P +an²/v²) (v - nb) = nRT
Where;
P = pressure (atm)
V = volume (L)
R = gas constant (0.0821 Latm/molK)
a and b = gas constant specific to each gas
T = temperature (K)
n = number of moles
According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol
Hence;
(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15
(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638
(P + 0.0291) (22.316) = 33.638
22.316P + 0.649 = 33.638
22.316P = 33.638 - 0.649
22.316P = 32.989
P = 32.989/22.316
P = 1.478
P = 1.48atm
