Answer:
Titrations. Because a noticeable pH change occurs near the equivalence point of acid-base titrations, an indicator can be used to signal the end of a titration. When selecting an indicator for acid-base titrations, choose an indicator whose pH range falls within the pH change of the reaction.
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Answer:
The correct answer is Pu, 234.
Explanation:
In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,
Mass of X = Mass of Molybdenum + Mass of Tin + Mass of neutrons
M = 1 * 103 + 1 * 131 + 2 * 0
M = 234
Now the atomic number (balanced) on both the sides,
Atomic number of X = Atomic number of Molybdenum + Atomic number of Tin + Atomic number of neutrons
A = 1*42 + 1*50 + 2*1
A = 94
The atomic number 94 is for the element Plutonium, whose symbol is Pu. Thus, the reactant is 234-Pu.
T = 14400 s
26.5 x 14400=381600 C
381600/96500=3.95 Faradays
Cu2+ + 2e- = Cu
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g
moles Au = 33.1 / 196.967 g/mol=0.168
Au+ + 1e- = Au
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday
0.168 x 96500=16217 Coulombs
16217 / 5.00=3243 s => 54 min
In gas the speed of sound is 343.2 meters per second, in liquid the speed of sound is 1,484 meters per second and in solids like steel it travels 5,120 meters per second.
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
