Answer:
a stable subatomic particle occurring in atomic nuclei,with a positive electric charge equal in magnitude to that of an electron.
Answer:
0.13M and 0.16m.
Explanation:
Molarity is defined as the moles of solute present in 1L of solution.
Molality is the moles of solute per kg of solvent.
To solve this problem we need to convert the mass of sucrose to moles using its molar mass and finding the volume in L of the solution and the mass in kg of solvent:
<em>Moles sucrose: </em>
Molar mass:
12C = 12*12.01g/mol = 144.12g/mol
22H = 22*1.005g/mol = 22.11g/mol
11O = 11*16g/mol = 176g/mol
Molar mass of sucrose is 144.12g/mol + 22.11g/mol + 176g/mol = 342.23g/mol
Moles are:
19.0g * (1mol / 342.23g/mol) = 0.0555 moles of sucrose
<em>Liters solution:</em>
425mL * (1L / 1000mL) = 0.425L
<em>kg solvent:</em>
425mL * (0.82g/mol) = 348.5g * (1kg / 1000g) = 0.3485kg
<em>Molarity:</em>
0.0555 moles / 0.425L
<h3>0.13M</h3><h3 />
<em>Molality:</em>
0.0555 moles / 0.3485kg
<h3>0.16m</h3>
.0002345 I believe this is correct
Answer:
c-do background research.
Explanation:
The scientific methods exposes the way scientists carry out their investigation and how findings are reported and at times discarded. It is systematic way of studying perceived observations in the environment.
During the course of an investigation, scientists may seek to find out about the prevailing knowledge about a phenomena and the level of research that might have been done in that regard. It is proper for such a scientist to do a background research by searching for related journals and publications in that field.
Answer:
C₇H₁₄O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.152 g
Mass of CO₂ = 2.726 g
Mass of H₂O = 1.116 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:
For carbon (C):
Mass of CO₂ = 2.726 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of C = 12/44 × 2.726
Mass of C = 0.743 g
For hydrogen (H):
Mass of H₂O = 1.116 g
Molar mass of H₂O = (2×1) + 16
= 2 + 16
= 18 g/mol
Mass of H = 2/18 × 1.116
Mass of H = 0.124 g
For oxygen (O):
Mass of compound = 1.152 g
Mass of C = 0.743 g
Mass of H = 0.124 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C) + (Mass of H)
Mass of O = 1.152 – (0.743 + 0.124)
Mass of O = 1.152 – 0.867
Mass of O = 0.285 g
Finally, we shall determine the empirical formula for the compound as follow:
C = 0.743 g
H = 0.124 g
O = 0.285 g
Divide by their molar mass
C = 0.743 / 12 = 0.062
H = 0.124 / 1 = 0.124
O = 0.285 / 16 = 0.018
Divide by the smallest
C = 0.062 / 0.018 = 3.44
H = 0.124 / 0.018 = 7
O = 0.018 / 0.018 = 1
Multiply by 2 to express in whole number.
C = 3.44 × 2 = 7
H = 7 × 2 = 14
O = 1 × 2 = 2
Empirical formula => C₇H₁₄O₂