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Elis [28]
4 years ago
9

A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca

lorimeter containing 300.0 grams of water, the temperature of the water increased from 20.0oc to 26.0oc. what is the specific heat of the unknown substance, assuming all the heat is transferred to the water?
Chemistry
1 answer:
MakcuM [25]4 years ago
3 0

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

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The iron nail displaces the Cu from the copper sulfate solution.  This results in the color of CuSO4's disappearance. Copper is less reactive than iron.

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Answer:

B

Explanation:

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3 years ago
Consider the following reversible reaction.
Marina86 [1]

Answer:

B is the correct option

Explanation:

K= the ratio of product of concentration of products to the product of concentration of reactants raised to power equal to their cofficients.

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The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g
lana66690 [7]

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

6 0
3 years ago
If the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, how high a column of water in meters can be suppor
dexar [7]

Answer:

The answer to this is

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

Explanation:

To solve this we first list out the variables thus

Density of the water = 1.00 g/mL =1000 kg/m³

density of mercury  = 13.6 g/mL = 13600 kg/m³

Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals

Therefore from the equation for denstity we have

Density = mass/volume

Pressure = Force/Area  and for  a column of water, pressure = Density × gravity×height

Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa

This value of pressure should be supported by the column of water as follows

Pressure = 101396.16 Pa =  kg/m³×9.81 m/s² ×h

∴  h = \frac{101396.16}{(1000)(9.81)} = 10.336 meters

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

4 0
3 years ago
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