Light comes from A. Nerve impulses. B. Electrons. C. An object vibrating. D. Photons

Schach [20]

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

8 0

3 years ago

Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha

elixir [45]

Answer:

0.269 g

Explanation:

$Po_{210} ^{84}$ ⟶ $Pb_{206} ^{82}$ + $+ He_{4} ^{2}$

Data:

Half-life of $Po_{210} ^{84}$ (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol

Time = 338.8 days

Isotopic masses

$Po_{210} ^{84}$ = 209.98 g/mol

$Pb_{206} ^{82}$ = 205.97 g/mol

Concepts

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.

The radioactive decay law is

N=Noe^(-λt)

Where:

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days

λ= radioactive decay constant

The radioactive constant is related to the half-life by the next equation

λ= $\frac{ln 2}{t(1/2)}$

so

λ= $\frac{ln2}{138.4 days}$ =0,005008 days^(-1)

No (Atoms of $Po_{210} ^{84}$ in t=0)

To get No we need to calculate the number of atoms of $Po_{210} ^{84}$ in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of $Po_{210} ^{84}$ in the initial sample.

This is:

$\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol}$ = 0,001599 mol of PoCl4

This is the number of mol of $Po_{210} ^{84}$ in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23

0,001599 mol of $Po_{210} ^{84}$ * 6.023*10^23 atoms/ mol of $Po_{210} ^{84}$ = 9.632 *10^20 atoms of $Po_{210} ^{84}$

Atoms after 338.8 days

We use the radioactive decay law to get this value

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of $Po_{210} ^{84}$ in the sample after 338.8 days has passed

The number of atoms $Po_{210} ^{84}$ transformed is equal to the number of atoms of $Pb_{206} ^{82}$ produced.

The number of atoms of $Po_{210} ^{84}$ transformed is No - N

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of $Pb_{206} ^{82}$ produced

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of $Pb_{206} ^{82}$ ) / ( 6.023*10^23 atoms of $Pb_{206} ^{82}$ / mol of $Pb_{206} ^{82}$ = 0.001306 moles of $Pb_{206} ^{82}$

This number of moles have a mass of:

(0,001306 moles of $Pb_{206} ^{82}$ )* (205.97 g of $Pb_{206} ^{82}$ /mol of $Pb_{206} ^{82}$ ) = 0.269 g

3 0

3 years ago

When studying pea plants, Gregor Mendel made many interesting discoveries and came to a few conclusions about what happens as a

lubasha [3.4K]

Characteristics of pea plants are determined by genes inherited from both the parents.

Option 4.

<h3><u>Explanation:</u></h3>

Gregor Johann Mendel is also called as the father of Genetics. Mendel experimented on the pea plants to determine what happens when different types of pea plants are crossed between themselves and what actually determines the characteristics of the pea plants. After two types of experiments, he came to a conclusion that there are certain substances which he named as '' factors'' which are actually inherited from the parents. He also told that there are certain type of factors which are dominant over the others which he termed as recessive factors. So, the characteristics of the progenies depend on both the factors inherited from both the parents.

6 0

3 years ago

Two major discoveries of the gold foil experiment

tiny-mole [99]

Rutherford's gold foil experiment proved that there was a small, dense, positively charged nucleus at the center, which contained most of the mass of the atom. Which contained electrons orbiting the nucleus.

5 0

3 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s

Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$