Answer:
Throughout the explanation section, the reason behind the given statement is described.
Explanation:
- The chemicals thus produced were indeed opposite or separate from the reaction mixture, this same reaction wouldn’t change when it's more balanced than some of the reactants.
- Another reason is that the development of advanced organisms, as well as chemical alterations, is irreversible throughout nature cant undo the chemical modifications.
Gloves is the tool component which will unlikely present a hazard in this scenario.
<h3>
What is Hazard?</h3>
This is referred to as potential source of harm when performing various types of activities.
Gloves will unlikely cause harm as it is used to prevent the risk of electric shock when handling electrical devices which is why it is the most appropriate choice.
Read more about Hazard here brainly.com/question/7310653
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The dissociation of formic acid is:
The acid dissociation constant of formic acid, 
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)
Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is 
.
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
<h3>Further explanation</h3>
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g
