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nordsb [41]
3 years ago
5

Calculate the volume of liquid in the tank sketched below. Give your answer in liters, and round to the nearest 0.1 L.

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

The answer is 18.9

Explanation:

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health issues and physical encounters

4 0
3 years ago
Combined gas law problem: a balloon is filled with 500.0 mL of helium at a temperature of 27 degrees Celsius and 755 mmHg. As th
Papessa [141]
The statement of the combined gas law for a fixed amount of gas is,
PV/T = constant
Here, the units of pressure and volume must be consistent and the temperature must be the absolute temperature (Kelvin or Rankine).
0.65 atm is equivalent to 494 mmHg
Using the equation:
(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)
V = 3396 ml = 3.4 liters
8 0
3 years ago
Let's just say I want to get my mixture to a certain pH but I add too much water to my solution. Can I just add the same volume
Dafna11 [192]

Answer:

You can

Explanation:

The concentration doesn't change when its equals out.

4 0
3 years ago
Ns
Brut [27]

Answer:

Ca(OH)₂ + H₂SO₄     →     CaSO₄ + H₂O

Explanation:

Chemical equation:

Ca(OH)₂ + H₂SO₄     →     CaSO₄ + H₂O

Balanced chemical equation:

Ca(OH)₂ + H₂SO₄     →     CaSO₄ + 2H₂O

The given reaction is double displacement reaction in which anion and cation of both reactant exchanged with each other. Calcium hydroxide react with sulfuric acid and form calcium sulfate and water.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AD +CB

6 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
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