Transition state <span> for the reaction between ethyl iodide and sodium acetateis shown below </span>
Answer:
773.43 torr
Explanation:
From the question given above, the following data were obtained:
Pressure (in inHg) = 30.45 inHg
Pressure (in torr) =?
We can convert 30.45 inHg to torr by doing the following:
1 inHg = 25.4 torr
Therefore,
30.45 inHg = 30.45 inHg × 25.4 torr / 1 inHg
30.45 inhg = 773.43 torr
Thus, 30.45 inhg is equivalent to 773.43 torr
Answer:
D. Protons ; neutrons.
Explanation:
For a given element the number of protons is fixed, but the number of neutrons differ because the element usually consist of a number of isotopes.
For example Carbon has isotopes which contain 12, 13 and 14 neutrons.
So that is why you average the number of neutrons to find the approximate atomic mass..
Answer:
1279 °C
Explanation:
Data Given:
Amount of Heat absorb = 5.82 x 10³ KJ
Convert KJ to J
1 KJ = 1000 J
5.82 x 10³ KJ = 5.82 x 10³ x 1000 = 5.82 x10⁶ J
mass of sample = 8.92 Kg
Convert Kg to g
1 kg = 1000 g
8.92 Kg = 8.92 x 1000 = 8920 g
Cs of steel = 0.51 J/g °C
change in temperature = ?
Solution:
Formula used
Q = Cs.m.ΔT
rearrange the above equation to calculate the mass of steel sample
ΔT = Q / Cs.m .... . . . . . (1)
Where:
Q = amount of heat
Cs = specific heat of steel = 0.51 J/g °C
m = mass
ΔT = Change in temperature
Put values in above equation 1
ΔT = 5.82 x10⁶ J / 0.51 (J/g °C) x 8920 g
ΔT = 5.82 x10⁶ J /4549.2 (J/°C)
ΔT = 1279 °C
So,
change in temperature = 1279 °C
Answer:
0.001 mole of NaF.
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 100 mL
Molarity = 0.01 M
Mole of NaF =?
Next, we shall convert 100 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 ml
100 mL = 0.1 L
Thus, 100 mL is equivalent to 0.1 L.
Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity = 0.01 M
Mole of NaF =?
Molarity =mole /Volume
0.01 = mole of NaF / 0.1
Cross multiply
Mole of NaF = 0.01 × 0.1
Mole of NaF = 0.001 mole.
Thus, 0.001 mole of NaF is present in 100 mL of the solution.
