A statement from the laboratory report which best represents a conclusion for this investigation is that: D. students who spend less time studying after school get lower math grade.
<h3>What is a research?</h3>
A research is also referred to a study and it can be defined as an investigation which typically involves the process of manipulating an independent variable (the cause), in order to determine or measure the dependent variable (the effect) and reach a logical conclusion with results.
<h3>What is an analytical research report?</h3>
An analytical research report can be defined as a well-crafted and written document that is typically used by a researcher to display a detailed analysis of the information which are obtained through a particular research method.
In this scenario, we can infer and logically deduce that a statement from the laboratory report which best represents a conclusion for this investigation is that the group of students who spend less time studying after school would get lower math grade.
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Explanation:
density(d) = mass(m) × volume(v)
given:
mass = 4.3 g = 0.0043 Kg
volume = 2.7 cm³ = 0.027 m³
= 0.0043 Kg / 0.0027 m³
= 0.15 Kg/m³
Answer: 0.53g
Explanation:
No of moles= volume ×molarity/1000
We have the volume and the molarity
Volume=4L
Molarity=1.7M
No of moles = 4×1.7/1000
No of moles= 0.0068moles
Remember also that
No of moles= mass given/molar mass
Molar mass of Al(OH)3
Al= 27
O=16
H=1
Molar mass = Al+(O+H)3
Molar mass= 27+(16+1)3
Molar mass= 27+(17)3
Molar mass = 27+51
Molar mass= 78g/mol
To get the mass
Mass given = no of moles × molar mass
Mass= 0.0068×78
Mass= 0.53g
The concentration of LiOH in the reaction is 0.103 M
<h3>What is the concentration of lithium hydroxide?</h3>
This is a neutralization reaction and the equation of the reaction is written as; H2SO4(aq) + 2LiOH(aq) ---->Li2SO4(aq) + H2O(l)
The average volume of acid used is obtained from;
8.54 + 8.51 + 8.68/ 3 = 8.58 mL
CA = 0.150 M
VA = 8.58 mL
CB = ?
VB= 25.0 mL
NA = 1
NB = 2
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB = CAVANB/VBNA
CB = 0.150 * 8.58 * 2/25.0 * 1
CB = 0.103 M
The concentration of LiOH in the reaction is 0.103 M
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