Answer:
Detail is given below.
Explanation:
Given data;
Abundance of Berkmarium-95 = 70%
Abundance of Berkmarium-97 = 28%
Abundance of Berkmarium-94 = 2%
Average atomic mass closer to which isotope = ?
Solution:
1st of all we will calculate the average atomic mass of Berkmarium.
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (70×95)+(28×97)+(2×94) /100
Average atomic mass = 6650 + 2716+ 188 / 100
Average atomic mass= 9554 / 100
Average atomic mass = 95.54 amu
The average atomic mass is closer to the isotope Berkmarium-95 because it is present in abundance as compared to the other two isotope. So this isotope constitute most of the part of Berkmarium.
A. the mass number
B. the period number
C. the atomic number
D. the name of the element
When a nucleus has too many protons, it can absorb an electron and create _________.
A. a proton
B. a neutron
C. an electron
D. an alpha ray
Which of the following occurs in beta decay?
A. A neutron of an atom emits a high energy proton.
B. A neutron of an atom emits a high energy electron.
C. An electron of an atom emits a high energy proton.
D. A neutron of an atom emits a second high energy neutron.
What kind of ray is given off after electron capture?
A. a beta ray
B. a delta ray
C. an alpha ray
D. a gamma ray
Answer:
Which of the following occurs in beta decay?
Answer: A neutron of an atom emits a high energy electron.
Explanation:
Via Educere/ Founder's Education
B) carbon dioxide (CO2), a common gas
C) water (H2O)
D) silicon carbide (SiC), a network solid
X(g) + 2Y(g) → 2Z(9)
When 3.0 mol of X and 3.0 mol of Y are placed inside a container with a volume of 1.0 dmº, they
react to form the maximum amount of Z.
The final temperature of the reaction vessel is 120°C.
What is the final pressure inside the reaction vessel?
А
4.49 x 10 Pa
B
9.80 x 106 Pa
C
1.47 x 10' Pa
D
1.96 x 10' Pa
Answer:
D
Explanation:
PV=nRT
n=nx + my
n=3mol+3mol
n=6mol
T=120+273=393K
P=nRT/V
P=6×0.0821×393/1
P=193atm
1atm=1.01325×10^5Pa
193atm=xPa
x=1.96×10^7Pa
The molar mass of NaBr is 102.9 grams and that of Pb(NO3)2 is 331.21 grams.
Answer: 193.4 grams
Explanation: According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number of particles.
To calculate the moles, we use the equation:
According to stoichiometry
1 mole of reacts with 2 moles of
0.94 moles of will react with= of
Mass of
Answer:
193.24g
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
Pb(NO3)2(aq) + 2NaBr(g) -> PbBr2(s) + 2NaNO3(aq)
Step 2:
Determination of the masses of Pb(NO3)2 and NaBr that reacted from the balanced equation. This is illustrated below:
Molar Mass of Pb(NO3)2 = 331.21g/mol
Molar Mass of NaBr = 102.9g/mol
Mass of NaBr from the balanced equation = 2 x 102.9 = 205.8g.
From the balanced equation,
Mass of Pb(NO3)2 that reacted = 331.21g
Mass of NaBr that reacted = 205.8g
Step 3:
Determination of the mass of NaBr that reacted with 311g of Pb(NO3)2. This is illustrated below:
From the balanced equation above,
331.21g of Pb(NO3)2 reacted with 205.8g of NaBr.
Therefore, 311g of Pb(NO3)2 will react with = (311x205.8)/331.21 = 193.24g of NaBr.
From the calculations made above, 193.24g of NaBr will react with 311g of Pb(NO3)2