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Sunny_sXe [5.5K]
3 years ago
14

Mention 4 products formed formed from destructive distillation of coal

Chemistry
2 answers:
rewona [7]3 years ago
7 0

Answer:

coal oil, coal gas, ammonia liquor, gas carbon.

LenaWriter [7]3 years ago
5 0

coal gas, gas carbon, coal tar and ammonia liquor,

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In Figure 13-3, which spheres have the GREATEST density (top, middle, or
natulia [17]

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bottom

Explanation:

the ones with the greatest density will sink. if they have a lighter density than water, they will float. if they have about the same, they will stay in the middle. and if they have a higher density they will sink.

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How many microliters are there in a volume of 370 mm 3of cobra venom?
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3700 microL is the answer

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A television set changes electrical energy to sound and light energy.In this process, some energy is
masha68 [24]

Answer:

lost

Explanation:

When a television set changes electrical energy to sound and light energy, some of the initial energy is lost in other forms usually heat.

Energy conversion or transformation from one form to another is not an 100% efficient process. This is compliance with the law of thermodynamics.

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3 years ago
Explain why samples of platinum and copper can have the same extensive properties but not the same intensive properties
JulsSmile [24]
<span>Extensive properties include mainly mass and volume. Platinum and copper can have the same extensive properties because depending on the type of sample size of each, you could get them to have the same volume and/or mass. However, since elements are given their intensive properties based on the number of protons they have, platinum and copper will never have the same intensive properties because each has a unique number of protons.</span>
3 0
3 years ago
A solution is prepared by adding 50.00 g of lactose (milk sugar) to 110.0 g of water at 55 °C. The partial pressure of water abo
Ede4ka [16]

Answer:

\boxed{\text{115.2 torr}}

Explanation:

Let’s call water Component 1 and lactose Component 2.

According to Raoult’s Law,  

p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

m₁ = 110.0 g;    p₁° = 118.0 torr

m₂ = 50.00 g; p₂° =    0    torr

1. Calculate the moles of each component

n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}

2. Calculate the mole fraction of each component

\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}

3. Calculate the vapour pressure of the mixture

p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is $\boxed{\textbf{115.2 torr}}$}

8 0
3 years ago
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