Answer:
The variables to be examined in relation to carbon dioxide use are the amount of light exposure and amount of dissolved CO2. Phenol red is yellow/orange under acidic conditions, that is when the pH of the solution is less than 7 (e.g. pH = 6). This occurs when the concentration of CO2 is high.
Explanation:
is this correct
Question is incomplete, complete question is;
A 34.8 mL solution of
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of
(aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

0.044 M is the molarity of
(aq).
0.33L more water is needed to dissolve it.
Cocaine is a stimulant substance obtained from the leaves of two Coca species native to South America, Erythroxylum coca and Erythroxylum novogranatense.
<h3>What is a coca?</h3>
Coca is a plant. The leaves of the coca plant are the source of cocaine, which is an illegal that is used nasally, injected, or smoked for mind-altering effects. Cocaine is also an FDA-approved Schedule II . This means cocaine can be prescribed by a healthcare provider, but the process is strictly regulated.
<h3>What happens when you smoke coca leaves?</h3>
The smoking of coca paste causes four distinct successive phases of mental disorder: euphoria, dysphoria, hallucinosis and paranoid psychosis. It can produce severe intoxication, prolonged or relapsing psychosis and, in some cases, end of life.
Learn more about cocaine here:
<h3>
brainly.com/question/944545</h3><h3 /><h3>#SPJ4</h3>
2HCL + __ Zn --> ZnCl2 + H2
i think that's right, not exactly sure.