Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
Answer:
38
Explanation:
Symbol: Sr
Atomic mass: 87.62 u
Electrons per shell: 2,8,18,8,2
Atomic number: 38
Electron configuration: [Kr] 5s2
Van der Waals radius: 255 pm
Valence electrons: two
The nurse<span> is </span>caring<span> for a </span>client<span> with a temperature of 104.5 degrees Fahrenheit. A health </span>care provider<span> prescribes 500 mg of an antibiotic intravenous While undergoing a soapsuds </span>enema, theclient reports abdominal<span> cramping.</span>
The balanced equation is
<span>2 C6H6 +15 O2 = 12 CO2 + 6 H2O </span>
<span>the ratio between C6H6 and CO2 is 2 : 12 </span>
<span>moles CO2 produced = 7.94 x 12 / 2 =47.6</span>
I'm guessing D or C, remember that the noble gases cannot combine