The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
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Argon = Ar
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Answer:
The answer is burning wood
Explanation:
Answer: C = 0.014M
Explanation:
From n= m/M= CV
m =43.5 M= 148, V=850ml
43.5/148= C× 0.85
C= 0.35M
Applying dilution formula
C1V1=C2V2
C1= 0.35, V1= 25ml, C2=?, V2= 600ml
0.35× 25 = C2× 600
C2= 0.014M
Answers:
(a) 30.55 °C
(b) 298 K and 77°F
(c) 204.44 °C and 477.44 K
(d) -320.8 °F and -196 °C
Explanation:
Converting °C into °F;
°F = °C × 1.8 + 32
Converting °F into °C;
°C = °F - 32 ÷ 1,8
Converting °C into K;
K = °C + 273
Converting K into °C;
°C = K - 273