Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by
q=m C (T2-T1) ........1
Given : q= 2.1200 J
the initial temperature of gold, T1 = 22.0Celcius
the final temperature of gold, T2 = 1064.4Celcius
specific heat of gold = 0.131
putting values in eq 1:
⇒ 2.1200 = m × 0.131 × (1064.4-22)
⇒ 2.1200 = m × 0.131 × 1042.4
⇒ 2.1200 / 136.5544
⇒ 0.01552494829
Since 1g= 0.01552494829 Pounds
Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
Learn more about temperature here: brainly.com/question/11464844
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Answer:
The first one is B, "Decreasing surface area."
Explanation:
This is because greater the surface area exposed, the more collisions that occur between the solvent and solute. I also just took the test myself and got it correct.
Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
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