Answer:
First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.
Jupiter, Saturn, Uranus, Neptune, Jovian... is your answer.
Answer: (2) decreasing the concentration of HCl(aq) to 0.1 M
Explanation: Rate of a reaction depends on following factors:
1. Size of the solute particles: If the reactant molecules are present in smaller size, surface of particles and decreasing the size increases the surface area of the solute particles. Hence, increasing the rate of a reaction.
2. Reactant concentration: The rate of the reaction is directly proportional to the concentration of reactants.
3. Temperature: Increasing the temperature increases the energy of the molecules and thus more molecules can react to give products and rate increases.
(1) Increasing the initial temperature to 25°C will increase the reaction rate.
(2) Decreasing the concentration of HCl(aq) to 0.1 M will decrease the reaction rate due to lesser concentration.
(3) Using 1.2 g of powdered Mg will increase the reaction rate due to large surface area.
(4) Using 2.4 g of Mg ribbon will increase the reaction rate due to high concentration of reactants.
To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.
n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2
At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:
V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L
Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
