Answer:
structure of a muscle cell
Answer:
Option 4 ) 1-butyne
Explanation:
In organic chemistry, you should use IUPAC convention in order to name an organic compound. First of all, you should identify the lenght of the organic chain, for this case, you have 5 C atoms, but in fact, you have a triple bond (which would be a substitute: ethynil-) as a substitute, so the main skeleton would have 4 C atoms (a butane)
Then, you start by numbering carbon N° 1 as the one that has the substitute (triple bound)-starting from the right, it would be the second C):
CH₃-CH₂-CH₂-C≡CH
Which will finally leads us to 1-butyne
Answer:
E. potassium (K) and bromine (Br)
Explanation:
An ionic bond is formed between compounds with a large electronegativity difference between them. It is usually between a metal and non-metal.
- Potassium is a true metal found in group 1 on the periodic table.
- Bromine is a highly electronegative non-metal which is a halogen.
- Potassium will lose one of its electrons which will be gained by the Bromine.
- The electrostatic attraction between the two species will cause the ionic bond to form.
- The ability of one specie willing to lose electron and the other gaining, is the main bed rock of ionic bonding.
There is nothing conserved<span> in this reaction. When writing a β </span>equation<span>, remember that in the nucleus, a neutron ( n ) decays into a proton ( p+ ) and a high energy electron which is known as the beta ( β ) particle. Because a new proton has formed, the atomic number of the original atom will increase by 1</span>
The new temperature of the piece of lead will be 37.5°C
SPECIFIC HEAT CAPACITY:
- The amount of heat absorbed or released by a substance can be calculated by using the following formula:
Q = m × c × ∆T
Where;
Q = quantity of heat absorbed or released (J)
m = mass of substance (g)
c = specific heat capacity
∆T = change in temperature (°C)
According to this question,
Q = 1.90J
m = 1.29g
T1 = 26°C
T2 = ?
c = 0.128 J/g°C
1.90 = 1.29 × 0.128 × (T2- 26°C)
1.90 = 0.165 (T2 - 26°C)
1.90 = 0.165T2 - 4.29
1.90 + 4.29 = 0.165T2
6.19 = 0.165T2
T2 = 37.5°C
Therefore, the new temperature of the piece of lead will be 37.5°C.
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