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3 years ago

Covert 0.2598kl to l

1 answer:

7 0

Kl in this problem means kilo liter. 1 kilo liter is equivalent to 1000 liter and equal to one cubic meter.

But in this problem, you are asking to convert this to liter.

So:

0.2598 kl = 1000 liter

--------------

1 kilo liter

So multiply, we get 259.8 liters or 260 liters.

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Which of the following is a compound machine?

il63 [147K]

Compound= many parts together therefore bike is the answer

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3 years ago

Read 2 more answers

A student weighed out a 2.055 g sample of a cobalt chloride hydrate, ConClmpH2O, where n, m, and p are integer values to be dete

topjm [15]

The given mass of cobalt chloride hydrate = 2.055 g

A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.

The mass of anhydrous cobalt chloride = 1.121 g anhydrate.

The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g

Converting mass of water of hydration present in the hydrate to moles using molar mass:

Mass of water = 0.934 g

Molar mass of water = 18.0 g/mol

Moles of water = $0.934 g * \frac{1 molH_{2}O }{18 g H_{2}O } =0.0519 mol H_{2}O$

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3 years ago

What's the charge on Ni in the compound NiCi3? *

ss7ja [257]

Answer:

Explanation:

Chlorine is anion with a -1 charge. But they are three chlorine atoms.

-1 * 3 = -3

So they have a -3 charge.

So to balance the compound, the nickel has to be a cation with a +3 charge.

-3 + 3 = 0

Furthermore, a chemical bond always has a 0 charge. Remember that.

Hope it helped! Rate my answer a 5 star if correct.

8 0

3 years ago

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl

nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of caffeine extracted is $P = 39.8 \ mg$

Explanation:

From the question above we are told that

The K value is $K = 4.6$

The mass of the caffeine is $m = 40 mg$

The volume of water is $V = 2 mL$

The volume of caffeine is $v_c = 2 mL$

The number of times the extraction was done is n = 3

Generally the mass of caffeine that will be extracted is

$P = m * [\frac{V}{K * v_c + V} ]^3$

substituting values

$P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3$

$P = 39.8 \ mg$

6 0

3 years ago

A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te

Grace [21]

Answer:

The calorimeter constant is = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal = 447 J/°C

5 0

3 years ago