<u>Answer:</u> A. constant
Without any air resistance, as the golf ball falls into the well, its speed is constant.
<u>Explanation:</u>
Given that when the ball is released,<em>there is no air resistance acting on ball. So the ball is freely falling.
</em>
The only force acting on the freely falling body is the gravitational force pulling the body towards the centre of the earth.Here the only acceleration acting on the body is the <em>acceleration due to gravity.</em>
The acceleration due to gravity is 
. so through out the journey of the ball,the acceleration is . <em>So as the acceleration is constant, the speed is also constant.</em>
The source below says 35.89g of NaCl dissolve in 100 g of water at 20 °C.
<span>(35.89g / 100 mL) x (2000 mL) = 718 g
Hope this helped </span>
Answer:
See explanation
Explanation:
2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)
Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles
Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles
2 moles of HCl reacts with 1 mole of calcium carbonate
x moles of HCl reacts with 0.00054 moles of calcium carbonate
x = 2 * 0.00054/1
x = 0.00108 moles of HCl
Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles
Reaction of HCl and NaOH
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Since the reaction is in the mole ratio of 1:1
0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH
Answer:
brainly
Explanation:
Adjust your day-to-day behaviors.
Replace your light bulbs.
Use smart power strips.
Install a programmable or smart thermostat.
Purchase energy efficient appliances.
Answer:
pH = 4.27. Porcentaje de disociación: 0.03%
Explanation:
El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
Donde la constante de equilibrio, Ka, es
Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]
Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:
[H⁺] = [X⁻]
[HX] es:
20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M
Reemplazando es Ka:
1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]
2.858x10⁻⁹ = [H⁺]²
5.35x10⁻⁵M = [H⁺]
pH = -log[H⁺]
<h3>pH = 4.27</h3>
El porcentaje de disociacion es [X⁻] / [HX] inicial * 100
Reemplazando
5.35x10⁻⁵M / 0.1732M * 100
<h3>0.03%</h3>
