Answer:
1-butanol has higher boiling point mainly due to presence of hydrogen bonding.
Explanation:
Diethyl ether is a polar aprotic molecule due to presence of polar C-O-C moiety. Hence only dipole-dipole intermolecular force exist between diethyl ether molecules.
1-butanol is a polar protic molecule due to presence of C-OH moiety. Therefore dipole-dipole force along with hydrogen bonding exist between 1-butanol molecules.
So, intermolecular force is higher in 1-butanol as compared to diethyl ether. Hence more temperature is required to break intermolecular forces of 1-butanol to boil as compared to diethyl ether.
So, 1-butanol has higher boiling point mainly due to presence of hydrogen bonding.
<em>The formula of calcium carbonate is CaCO3
</em>
<em>The formula of nitric acid is HNO3.
</em>
<em>When put together:</em>
<em>CaCO3 + HNO3 </em><em>= </em><em>Ca(NO3)2 + CO2 + H2O
</em>
<em>
The balanced equation:</em>
<em>
CaCO3 + 2HNO3</em><em> = </em><em>Ca(NO3)2 + CO2 + H2O
</em>
<em>
</em>
<em></em>
Answer:
The question is incorrect and incomplete. Here's the correct question:
It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy when burned. To illustrate this difficulty,a) calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 23.5 °C to 100 °C , it boils, and the resulting steam is raised to 315 °C. b)Discuss additional complications caused by the fact that crude oil has less density than water.
Explanation:
Q= mc ΔT
Q= heat energy
m is mass
ΔT is change in temperature and c is specific heat capacity
calculating heat for latent heat of vaporisation
Q= ml where l is latent heat of vaporisation
a) Total heat energy used= heat required to raise temperature from 23.5 °C to 100 °C, heat required to boil water and heat required to further raise temperature from 100 °C to 315°C
Q = mc ΔT₁ + mL + mc ΔT₂
Q = m(c ΔT₁ + L + c ΔT₂)
m= Q÷(c ΔT₁ + L + c ΔT₂)
Q= 2.8 X 10⁷ J
c=4186J/kg°C
L=2256 x 10³J/kg
ΔT₁=76.5°C(100°C-23.5°C)
ΔT₂= 215°C(315°C-100°C)
(c ΔT₁ + L + c ΔT₂)= 4186J/kg°C *76.5°C + 2256 x 10³J/kg + 4186J/kg°C*215°C =3476219J/Kg
m= 2.8 x 10⁷J ÷3476219J/Kg
m =80.54 Kg
volume = mass÷ density
=80.54kg ÷ 10³kg/m³( density of water)
=0.0854m³
0.001m³ = 1 lL0.08054m³= 0.08054m³ /0.001m³= 80.54L
VOLUME is 80.54litres
b) since the density of crude is less than the density of water,and 80L of additional water is added, it'll make the crude to float on water thus inhibiting the extinguishing process
Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
