Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
The correct answer for the question that is being presented above is this one: "metal." <span>A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metal. This metal is a good testing for conductivity of an electricity but cannot be pulled into wires.</span>
<span> A </span>mixture<span> is made from </span>two<span> or more substances that are chemically unlike</span><span> and are not chemically joined. A </span>compound<span> is a substance formed when </span>two<span> or more elements chemically react with each other to ... substances because no new substance is formed, therefore they do not </span>have<span> any fixed properties.</span>
A pi bond... hope this helps!!!!!
