D = M/V = 76g / 22ml = 3.4g/ml
Half ~ D = 38g / 11ml = 3.4g/ml
Even if the object you had was cut in half, it’s density would remain the same.
The chemical equation is
Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)
Answer:
12
Explanation:
In the right hand side of the equation, there are three compound which contains O2, which are;
Cu(NO3)2 , number of oxygen atoms =3*2 =6
2NO2, number of oxygen atoms = 2*2=4
2H2O, number of oxygen atoms =2*1=2
Total number of oxygen atoms on the right side of equation = 6+4+2 =12
<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.
Answer:
The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,
Explanation:
To answer this problem we need to keep in mind the <u>definition of pH</u>:
So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:
-5.2 = log [H⁺]
= [H⁺]
6.31 x 10⁻⁶ M = [H⁺]
-5.6 = log [H⁺]
= [H⁺]
2.51 x 10⁻⁶ M = [H⁺]