Answer:
52 amu
Explanation:
To get the relative atomic mass of the element, we need to take into consideration, the atomic masses of the different isotopes and their relative abundances. We simply multiply the percentages with the masses. This can be obtained as follows:
[89/100 * 52] + [8/100 * 49] + [3/100 * 50]
46.28 + 3.92 + 1.5 =51.7 amu
The approximate atomic mass of element x is 52 amu
Answer:- 64015 J
Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.
density of water is 1 g per mL.
So, the mass of water =
= 4250 g
Final temperature of water after adding the hot copper bar to it is 26.15 degree C.
So,
for water = 26.15 - 22.55 = 3.60 degree C
Specific heat for water is 4.184 
The heat gained by water is calculated by using the formula:

where, q is the heat energy, m is mass and c is specific heat.
Let's plug in the values in the formula and do the calculations:

q = 64015 J
So, 64015 J of heat is gained by the water.
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 =
or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
Answer:
2.4 ×10^24 molecules of the herbicide.
Explanation:
We must first obtain the molar mass of the compound as follows;
C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1
We know that one mole of a compound contains the Avogadro's number of molecules.
Hence;
169 g of the herbicide contains 6.02×10^23 molecules
Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.
Answer:
100g/mol
Explanation:
Given parameters:
Mass of unknown gas = 2g
Volume of gas in flask = 500mL = 0.5dm³
Unknown:
Molar mass of gas = ?
Solution:
Since we know the gas is at STP;
1 mole of substance occupies 22.4dm³ of space at STP
Therefore,
0.5dm³ will have 0.02mole at STP
Now;
Number of moles =
Molar mass =
=
= 100g/mol