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2 years ago

The picture above represents which of the following?

Chemistry

1 answer:

PSYCHO15rus [73]2 years ago

6 0

This is a plant cell because there are chloroplasts in this diagram which are present in plant cells

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Find the boiling point temperature at 760 torr of an isomer of octane, c8h18, if its enthalpy of vaporization is 38,210 j mol-1

mixas84 [53]

According to Clausius-Clayperon equation,

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

$P_{1}$ is the vapor pressure at boiling point = 760 torr

$P_{2}$ is the vapor pressure at T_{2} =638.43 torr

Temperature $T_{2} = 110.0^{0}C + 273 = 383 K$

Δ $H_{vap} = 38210 J/mol$

Plugging in the values, we get

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

ln $\frac{760 torr}{638.43 torr} =\frac{38210 J/mol}{8.314 J/(mol.K)} (\frac{1}{383 K}-\frac{1}{T_{1} }$

$T_{1} = 389 K$

Therefore, the boiling point of octane = 389 K - 273 = $116^{0}C$

7 0

2 years ago

2) A solution consists of 0.50 mole of CaCl2 dissolved i

Aleksandr-060686 [28]

A. a higher boiling point and a lower freezing point

Explanation:

hope this helps :)

6 0

2 years ago

Read 2 more answers

375ml of a 0.455m sodium chloride solution is dilluted with 1.88l of water. what is the new concentration in molarity

mestny [16]

0.375 L of the NaCl solution contains 0.455 moles (1)

since it was diluted to 1.88 L

then let 1.88 L of solution contain x moles (2)

by combining both (1) and (2)

⇒ (0.345 L) x = (0.455 mol) (1.88 L)

⇒ x = (0.242 mol · L)  ÷ (0.345 L)

⇒ x = 0.702 mol


Thus the molarity of the NaCl after being diluted is 0.702 M OR 0.702 mol / L OR 0.702 mol/dm³.

7 0

3 years ago

Read 2 more answers

PLEASE HELP WITH THIS ASAP

iragen [17]

Radio waves
microwaves
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4 0

2 years ago

2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti

olga2289 [7]

Answer:

$\large \boxed{\text{933 J}}$

Explanation:

There are three heat transfers involved.

heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0

q₁ + q₂ + q₃ = 0

m₁ΔH + m₂C₂ΔT + C_calΔT = 0

Data:

m₁ = 2.1 g

m₂ = 280 g

Ti = 25.00 °C

T_f = 26.55 °C

Ccal = 92.3 J·°C⁻¹

Calculations:

Let's calculate the heats separately.

1. q₁

q₁ = 2.1 g × ΔH = 2.1ΔH g

2. q₂

ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C

q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J

3. q₃

q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J

4. ΔH

$\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}$

7 0

2 years ago