The empirical formula of arginine is C3H7N2O2.
<h3>What is the empirical formula of arginine?</h3>
Now we know that the empirical formula is the formula that shows the ratio of the atoms present in the compound.
Now we know that;
C - 41.368/12 H - 8.101/1 N - 32.162/14 O - 18.369/16
C - 3.445 H - 8.101 N - 2.297 O - 1.148
Dividing through by the lowest ratio;
C - 3.445/1.148 H - 8.101/1.148 N - 2.297/1.148 O - 1.148/1.148
C - 3 H - 7 N - 2 O - 2
C3H7N2O2
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Answer:
46.004 g/mol
Explanation:
Scottish physicist Thomas Graham formulated a law known as Graham's law of effusion in 1848. He conducted an experiment and found the relationship between the rate of effusion of a gas and its molar mass as:
where,
r is the rate of effusion of a gas
M is the molar mass of the gas.
Also, r = v/t
And for two gases taking different t₁ and t₂ to effuse, the formula is:
So,
For Neon :
= 8.61 minutes
= 20.1797 g/mol
For unkown gas:
= 13.0 minutes
= ? g/mol
<u>Molar mass of unknown gas = 46.004 g/mol</u>
Answer:
1 × 10-¹ moldm-³
Explanation:
Pkw = PH + POH
14 = 13 + POH
POH = 1
so , [OH-] = 1 × 10-¹ moldm-³
A + BC ---> AB + C
So here one reactant (A) is accepting a group which is being given by another compound (BC) however the A is not giving any group / element or ion
So this single displacement
Similarly in the given reaction
the anion OH- is only being replaced
The element Ca accepts OH- and H2O loses the same group to form new element H2
So the correct answer is
Single replacement also known as single displacement.
Simply its the smallest particle da