I don’t know what plant were presented in your lesson. So, I did sunflowers. The tropism I used was photo tropism
Answer:
C
Explanation:
C. Nucleophilic attack and carbocation rearrangement to form the most stable carbocation before the substitution reaction
Sn(s) + 2 HF(aq) ----> SnF2(s) + H2(g)
M HF = 20 g/mol
M SnF2 = 157 g/mol
40 g (HF) -----> 157 g (SnF2)
30 g (HF) -----> x g (SnF2)
x = 30 g × 157 g / 40 g = 117,75g
Answer: 117,75 g
:-) ;-)
Data:
V (volume) = 500.0 mL = 0.5 L
T (temperature) = 15.00ºC
(converting in Kelvin) → TK = TC + 273 → TK = 15 + 273 = 288 K
P (pressure) = 736.0 mmHg
R (constant) = 62.363 (mmHg*L/mol*K)
m (mass) = 2.688 g
M (Molar Mass) = ? (g/mol)
Formula: General Gas Equation
Solving:
Product of extremes equals product of means:
Therefore: <span>The gas found to have such a molar mass is
xenon gas</span>
Answer:
The percent composition is 21% N, 6% H, 24% S and 49% O.
Explanation:
1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.
In 1 mole of (NH4)2SO4, there are:
- 2 moles of N.
- 8 moles of H.
- 1 mole of S.
- 4 moles of O.
2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:
- 2 moles of N (14g/mol) = 28g
- 8 moles of H (1g/mol) = 8g
- 1 mole of S (32g/mol) = 32g
- 4 moles of O (16g/mol) = 64g
3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:
In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.
So, the percent composition is 21% N, 6% H, 24% S and 49% O.