Answer:
58.94 mL
Explanation:
V1 = 48.3 mL V2 = v mL
T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k
We will use the gas equation:
PV = nRT
Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,
we can say that
V / T = k , (where k is a constant)
Since this is the first case,
V1 / T1 = k --------------------(1)
For case 2:
Since we have the same constants, the equation will be the same
V / T = k (where k is the same constant from before)
V2 / T2 = k (Since this is the second case) ------------------(2)
From (1) and (2):
V1 / T1 = V2 / T2
Now, replacing the variables with the given values
48.3 / 295 = v / 360
v = 48.3*360 / 295
v = 58.94 mL
Therefore, the final volume of the gas is 58.94 mL
Explanation:
Both have the same temperature, but the large bowl of soup has more thermal energy than the small mug because there is more matter present in the bowl.
CTSO is an extracurricular program for students. The program helps the students build leadership qualities and career-specific skills.
<h3>What is CTSO?</h3>
A career and technical student organization is a program designed and implemented for the students to improve their interpersonal relationships, goal setting, and build leadership skills.
CTSO helps in building the skills required for professional career establishment and for the educational purposes that are other than academics. It includes competitions, plays, sports events, etc.
Therefore, CTSO increases leadership skills and provides career-specific skills.
Learn more about CTSOs here:
brainly.com/question/11454436
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I believe the correct answers from the choices listed above are options two and five. It would be metals and hydroxyl ions that are common to bases. <span>Any of a class of compounds that form hydroxyl ions (OH) when dissolved in water, and whose aqueous solutions react with acids to form salts. Hope this answers the question.</span>
Answer:
Explanation:
Formula of percent composition:
% X = ( mass of element in compound / total mass of compound ) × 100
11) Al₂O₃
Al = 27 g/mol
O = 16 g/mol
%Al = (54 / 101.96) × 100
%Al = 52.9%
%O = (48 / 101.96) × 100
%O = 47.1%
12) C₅H₅O₅
C = 12 g/mol
H = 1 g/mol
O = 16 g/mol
C% = (60 / 145 ) × 100
C% =41.4%
H% = (5 / 145 ) × 100
H% =3.4%
O% = (80 / 145 ) × 100
O% = 55.2%
13) BaBr₂
Ba = 137.3 g/mol
Br =79.9 g/mol
Ba% = (137.3 / 297.14) × 100
Ba% = 46.2%
Br% = (159.8/ 297.14) × 100
Br% = 53.8%
14) Fe₃(PO₃)₂
Fe = 55.8 g/mol
P = 31 g/mol
O = 16 g/mol
Fe = (167.4/325.48) × 100
Fe = 51.43%
P = (62/325.48) × 100
P = 19.05%
O = (96 /325.48) × 100
O = 29.49%