Answer:
Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.
Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
Explanation:
Initial Pressure = 24 lb in-2
Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)
Final Pressure = ?
Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)
No Change in Volume.
From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P1T1 = P2T2
P2 = P1T1 / T2
P2 = 24 * 268 / 308 = 20.88 lb in-2
There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.
Answer:
pure water, pH = 7.0 (Neutral)
lake water, pH = 6.5 (Acidic)
baking soda solution, pH = 9 (Alkaline)
soapy water, pH = 12 (Alkaline)
Explanation:
The degree of acidicity or alkalinity of a solution can be determined on a pH meter. A pH below 7 is acidic; a pH of 7 is neutral; a pH value of above 7 is alkaline.
Answer:
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
Explanation:
Data Given:
Moles = n = 3.2 mol
Temperature = T = 312 K
Pressure = P = ?
Volume = V = 87 m³ = 87000 L
Formula Used:
Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for P,
P = n R T / V
Putting Values,
P = (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
