The other students in the lab should be notified next in this type of scenario.
<h3>What is an acid?</h3>
This is a substance which donates protons and is very corrosive. It also turns blue litmus paper red.
When it was spilled and baking soda was used to neutralize it on the floor , it is best to inform the other students too so as to prevent them from being exposed by mistake thereby reducing risk of injury.
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Answer:
0.0344 moles and 1.93g.
Explanation:
Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:
<em>Moles KOH:</em>
15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles
<em>Mass KOH:</em>
0.0344 moles * (56.11g/mol) = 1.93g of KOH
Answer:
A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time
Explanation:
this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible
Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>
