Answer:
<em><u>700N</u></em>
Explanation:
<em><u>F</u></em><em><u>=</u></em><em><u>ma</u></em>
<em><u>mass</u></em><em><u>=</u></em><em><u>35</u></em>
<em><u>acceleration</u></em><em><u> </u></em><em><u>=</u></em><em><u>20</u></em>
Answer:
31.24 kJ
Explanation:
- SiO₂(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol
First we <u>convert 3.00 grams of SiO₂ to moles</u>, using its <em>molar mass</em>:
- 3.00 g SiO₂ ÷ 60.08 g/mol = 0.05 mol
Now we <u>calculate the heat absorbed</u>, using the <em>given ΔH°</em>:
If the complete reaction of 1 mol of SiO₂ absorbs 624.7 kJ, then with 0.05 mol:
- 0.05 mol * 624.7 kJ/mol = 31.24 kJ of heat would be absorbed.
Answer:
0.125. work- divide the volume value by 1000
873. work- multiply the length value by 100
98100. work- Conversion factor: 1 kg = 100000 cg
1) Centigram = Kilogram * 100000
2) Centigram = 0.981 * 100000
3) Centigram = 98100
285.65. work- 12.5°C + 273.15 = 285.65K
446.85. work- 720K − 273.15 = 446.85°C
346.25. work- 73.1°C + 273.15 = 346.25K
Explanation:
i hope this helps:) brainliest plss??
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Acids: taste sour, has a pH less than 7
bases: tastes bitter, has a pH greater than 7
