Answer:
In advance
middle
lower
Explanation:
These are the safety precautions needed when carrying out duties in the fume hood.
When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal <u>in advance</u> to reduce and minimize the raising and lowering of the hood sash at intervals.
It is also pertinent to understand that working in the<u> middle </u>of the work surface helps to promote the movement of air and keeps the area neat and tidy.
However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to <u>lower </u>the sash at that point in time.
Answers are:
Catabolism:
- g<span>enerally exergonic (spontaneous): In this reactions energy is released.
- </span><span>convert NAD+ to NADH. Electrons and protons released in reactions are attached to NAD+.
- </span><span>generation of ATP. ATP is synthesis from ADP.
- </span><span>convert large compounds to smaller compounds. Foe example starch to monosaccaharides.
Anabolism:
</span><span>- convert NADPH to NADP+. Protons and electrons are used to make chemical bonds.
</span>- <span>convert small compounds to larger compounds.</span>
Answer:
The hydrogen ion concentration of the substance is 3.16*10⁻⁵ M
Explanation:
pH is a parameter used to measure the degree of acidity or alkalinity of a substance. The pH is calculated as the negative logarithm (base 10) of the concentration of hydronium ions [H₃O⁺] or hydrogen ions [H⁺].
pH= - log [H₃O⁺]= - log [H⁺]
Values on the pH scale range from 0 to 14, where pH equal to 7 is neutral, below 7 is acidic and above 7 is basic.
In this case, pH= 4.5
So:
4.5= - log [H⁺]
Solving:
[ H⁺]= 10⁻⁴ ⁵
[H⁺]= 3.16*10⁻⁵ M
The hydrogen ion concentration of the substance is 3.16*10⁻⁵ M
Answer:
Explanation:
H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄
29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄
Let the concentration of H ₂ S O ₄ be S₂ .
In terms of normal or equivalent solution is will be 2 N solution
From the formula S₁ V₁ = S₂ V₂
= 29.09 x .639 = 213.8 x S₂
S₂ = .087 N solution
In terms of molar solution it will be .087 / 2 M
= .0435 M