Answer:
Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:
The volume of the gas starts at 250 milliliters and the temperature is 137 °C.
The volume of the gas is increased to 425 milliliters, but the temperature is unknown.
We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.
Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.
The units of milliliters (mL) cancel.
The temperature changes to <u>232.9 degrees Celsius.</u>
Rutherford was one of the early scientists who worked on the atomic model. Before his discovery of the nucleus, the widely accepted theory was J.J Thomson's Plum Pudding Model. In this model, all the protons, electrons and neutrons are in the nucleus. But the electrons are more in number such that the electrons act as the 'pudding' and the proton and nucleus the 'plum'. This was Rutherford's hypothesis in his gold foil experiment. In order to test the Plum Pudding model, he hypothesized that when a beam of light is aimed at the atom, it would not diffract because the charges in the nucleus are well-distributed. However, his experiment disproved Thomson's model. Some light indeed passed through but a few was diffracted back to the source. He concluded that this was because there is a dense mass inside the atom called nucleus. Thus, from there on, he proposed the model that the electrons are orbiting around the nucleus.
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
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