Ok i help you answer these question
Answer:
The cyanidin indicator turns blue within a pH range of 5 - 7. The pH of the solution could be 5, 6 or 7.
An indicator is used to determine the endpoint of a titration.
Explanation:
Cyanidin indicator changes colour with each change in pH. In acidic solutions (pH < 7) cyanidin indicator will turn red, through to purple and blue, while in basic solutions (pH > 7), cyanidin indicator will change colour from aquamarine through to green and yellow. The cyanidin indicator turns blue within a pH range of 5 - 7.
Titration is a technique used in analytical chemistry to determine the unknown concentration of a solution. A solution of known concentration is added from a burette to the solution of unknown concentration until the reaction between the two solutions is complete. This known as the endpoint of the experiment. The endpoint of a titration is determined using an indicator which is added to reaction mixture. A colour charge is produced by the indicator at the endpoint of the reaction.
Note: An indicator is a dye of weak organic acids or bases which changes colour with changes in the pH of a solution. Some common indicators are methyl orange, methyl red, phenolphthalein, etc. These indicators are used to monitor the changes in the pH of solutions during a reaction.
Explanation:
<em>According</em><em> </em><em>to</em><em> </em><em>your</em><em> </em><em>question</em><em>, </em>
<em>no</em><em>.</em><em> </em><em>a</em><em>.</em><em> </em><em>ans</em><em> </em><em>would</em><em> </em><em>be</em><em> </em><em>like</em><em>;</em><em> </em><em>chemical</em><em> </em><em>formula</em><em> </em><em>is</em><em> </em><em>defined</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>an</em><em> </em><em>expression</em><em> </em><em>which</em><em> </em><em>determines</em><em> </em><em>no</em><em>.</em><em> </em><em>and</em><em> </em><em>type</em><em> </em><em>of</em><em> </em><em>molecule</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>compound</em><em>. </em>
<em>b</em><em>.</em><em> </em><em>no</em><em>.</em><em> </em><em>ans</em><em>;</em><em> </em><em>it</em><em> </em><em>tells</em><em> </em><em>that</em><em> </em><em>what</em><em> </em><em>type</em><em> </em><em>of</em><em> </em><em>compound</em><em> </em><em>is</em><em> </em><em>formed</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>type</em><em> </em><em>and</em><em> </em><em>no</em><em>.</em><em> </em><em>of</em><em> </em><em>atoms</em><em> </em><em>present</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>atom</em><em>.</em>
<em>c</em><em>.</em><em> </em><em>no</em><em> </em><em> </em><em>ans</em><em>;</em><em> </em><em>the</em><em> </em><em>formulation</em><em> </em><em>of</em><em> </em><em>h2so4</em><em> </em><em>states</em><em> </em><em>that</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>acid</em><em> </em><em>named</em><em> </em><em>as</em><em> </em><em>hydrochloric</em><em> </em><em>acid</em><em> </em><em>which</em><em> </em><em>is</em><em> </em><em>formed</em><em> </em><em>by</em><em> </em><em>reacting</em><em> </em><em>of</em><em> </em><em>hydrogen</em><em> </em><em>(</em><em>2</em><em> </em><em>atoms</em><em> </em><em>)</em><em> </em><em>,</em><em>sulpher</em><em> </em><em>(</em><em>*</em><em>1</em><em>atom</em><em>)</em><em> </em><em>and</em><em> </em><em>oxygen</em><em>(</em><em>4</em><em>atoms</em><em>)</em><em>.</em>
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
The cool thing about chemical equations is that the mole ratios between compounds is the same as the ratios shown in the balanced equation.
So first, check to make sure the equation is balanced. I just looked at it, and it is (if you can't tell, that means there're the same number of each kind of atom on both sides of the equation).
The answer lies in the coefficients.
So, you'll notice there's a 3 in front of the sodium hydroxide and no coefficient (which implies 1) in front of the iron trichloride.
It's literally that simple: 3:1.
The answer is D.