Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.
Answer:
4.92 L
Explanation:
Rearrange ideal gas law and solve.
Change C to K.
- Hope that helps! Please let me know if you need further explanation.
Answer:
28.43 min
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given that:
The rate constant, k =
min⁻¹
Initial concentration
= 0.1 M
Final concentration
=
M
Time = ?
Applying in the above equation, we get that:-




For the answer to the question above asking w<span>hen an atom of n-14 is bombarded by an alpha particle, the single product is?
</span> <span>You're starting with 14/7 N, correct?
An alpha particle is two protons, two neutrons, which is 4/2, correct?
</span><span>So I</span> think the answer to your question is the third one which is <span>c. 18/9 f </span>