Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Answer:
113.8g
Explanation:
Statement of problem: mass of 1.946mole of NaCl
Given parameters:
Number of moles of NaCl = 1.946mole
Unknown: mass of NaCl
Solution
To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:
Number of moles = 
To find the molar mass of NaCl:
the atomic mass of Na = 23g
atomic mass of Cl = 35.5g
Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹
Mass of NaCl = Number of moles x molar mass of NaCl
Mass of NaCl = 1.946 x 58.5 = 113.8g
Answer:
m = 20.9 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:
Regards!
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
Adding perchloric acid to water would cause it's conductivity to increase.
