Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :
Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
The period is the end of the sentence!!!
Answer:
Observing. This is the most basic skill in science. ...
Communicating. It is important to be able to share our experiences. ...
Classifying. After making observations it is important to notice similarities, differences, and group objects according to a purpose. ...
Inferring. ...
Measuring. ...
Predicting.Explanation:
hope this he;lp
pick me a the brainliest
Answer:
∇T = 51.68°C
Explanation:
Mass = 150g
Heat Energy (Q) = 1.0*10³J
Change in temperature ∇T = ?
Q = mc∇T
Q = heat energy
M = mass
C = specific heat capacity of the gold = 0.129j/g°C
∇T = change in temperature
Q = Mc∇T
1.0*10³ = 150 * 0.129 * ∇T
1000 = 19.35∇T
Solve for ∇T
∇T = 1000 / 19.35
∇T = 51.679°C = 51.68°C
The change in temperature of gold was 51.68°C
