Answer:
4. 1.18 mol·L⁻¹
14. See below.
Explanation:
4. Dilution calculation
V₁c₁ = V₂c₂
Data:
V₁ = 200 mL; c₁ = 5.6 mol·L⁻¹
V₂ = 950 mL; c₂ = ?
Calculation:
c₂ = c₁ × V₁/V₂
c₂ = 5.6 mol·L⁻¹ × (200/950) = 1.18 mol·L⁻¹
The new concentration is 1.18 mol·L⁻¹
.
14. Boyle's Law graphs
We can write Boyle's Law as
pV = k or p = k/V or V= k/p
p and V are inversely related.
(a) As pressure increases, volume decreases. Thus, a graph of V vs p is a hyperbola.
(b) p = k/V =k(1/V)
1/V = (1/k)p
y = m x + 0
A graph of 1/V vs p is a straight line.
Answer:
La respuesta a tu pregunta esta abajo.
Explanation:
Cuando un barco navega en un río en el sentido de la corriente, la velocidad que lleva más el impulso que le proporciona la corriente, hace que avance más rápido llegando más pronto a su destino sin tanto esfuerzo del motor. Por lo que gastaría una menor cantidad de combustible.
Cuando el barco se desplaza en sentido contrario a la corriente del motor, la corriente del río hace que su desplazamiento neto sea menor ya que el barco avanza y el río lo regresa un tramo y el esfuerzo del motor el mayor.
En estas condiciones aunque el barco tenga que recorrer la misma distancia que en el caso anterior, en realidad debería sumarse la distancia que lo regresa la corriente y por lo tanto el combustible gastado sera mayor.
Answer:
40.95 L
Explanation:
We'll begin by calculating the number of mole in 12.3 g of O₂. This can be obtained as follow:
Mass of O₂ = 12.3 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mole of O₂ =?
Mole = mass / Molar mass
Mole of O₂ = 12.3 / 32
Mole of O₂ = 0.384 mole
Next, we shall determine the volume occupied by 0.384 mole of O₂. This can be obtained as follow:
Number of mole (n) of O₂ = 0.384 mole
Pressure (P) = 1 atm
Temperature (T) = 273 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of O₂ =?
PV = nRT
1 × V = 0.384 × 0.0821 × 273
V = 0.384 × 0.0821 × 273
V = 8.6 L
Thus, the volume of O₂ is 8.6 L
Finally, we shall determine the volume of air that contains 8.6 L of O₂. This can be obtained as follow:
Volume of O₂ = 8.6 L
Percentage of O₂ in air = 21%
Volume of air =?
Percentage of O₂ = Vol of O₂ / Vol of air × 100
21% = 8.6 / Vol of air
21 / 100 = 8.6 / Vol of air
Cross multiply
21 × Vol of air = 100 × 8.6
21 × Vol of air = 860
Divide both side by 21
Volume of air = 860 / 21
Volume of air = 40.95 L
Therefore, the volume of air is 40.95 L.
Atomic radius, potassium is farther down on the table
Answer:
Option D. The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.
Explanation:
Solubilities of solutes are enhanced when the temperature is increased.
From the experiment conducted,
It is evident that glass B temperature is higher than glass A temperature, because the solute dissolves faster in glass B than in glass A . This implies that glass A is cooler than glass B, hence the particles in A will move slower than that in B.
