Within the core of the Sun, temperatures and pressures are high enough to fuse hydrogen atoms into helium, which is the Sun's main form of energy production. Assuming there was a slight mistake in where you have copied the results here the correct answer is the third option.
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The balanced equation :
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
<h3>Further explanation</h3>
Given
Reaction
NaHCO(s) --> _CO2+_NaCO(s)+_H2O
Required
The balanced equation
Solution
Maybe the equation should be like this :
NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Give a coefficient
NaHCO₃⇒aCO₂ + bNa₂CO₃+cH₂O
Make an equation
Na, left=1, right=2b⇒2b=1⇒b=1/2
H, left=1, right=2c⇒2c=1⇒c=1/2
C, left=1, right=a+b⇒a+b=1⇒a+1/2=1⇒a=1/2
The equation becomes :
NaHCO₃⇒1/2CO₂ +1/2Na₂CO₃+1/2H₂O x2
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:
A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4
As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.
Answer: The amount of carbon-14 left after 10 years is 25 g
Explanation:
Formula used :

where,
a = amount of reactant left after n-half lives = ?
= Initial amount of the reactant = 100 g
n = number of half lives =
Putting values in above equation, we get:


Therefore, the amount of carbon-14 left after 10 years is 25 g
Answer:
c.
Explanation:
it is c.-1 because oxidation number of cl is-1