Answer:
answer is a since solute dissolve a solvent to give a solution
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
Answer:
Option C :
a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio
Explanation:
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
Tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
For Example
C₂H₆O₂ Consist of Carbon (C), Hydrogen (H), and Oxygen (O)
Now
Look at the ratio of these three atoms in the compound
C : H : O
2 : 6 : 2
Divide the ratio by two to get simplest ratio
C : H : O
2/2 : 6/2 : 2/2
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₂H₆O₂ = CH₃O
So, Option C is correct :
a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
Answer:
Iron remains = 17.49 mg
Explanation:
Half life of iron -55 = 2.737 years (Source)
Where, k is rate constant
So,
The rate constant, k = 0.2533 year⁻¹
Time = 2.41 years
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
= 32.2 mg
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
is the initial concentration
So,
![[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.2533%5Ctimes%202.41%7D%5C%20mg)
![[A_t]=32.2\times e^{-0.610453}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.610453%7D%5C%20mg)
![[A_t]=17.49\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D17.49%5C%20mg)
<u>Iron remains = 17.49 mg</u>
