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3 years ago

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2. A flask and stopper have a mass of 110.525 g. A 50.0-mL sample of gasoline is transferred to the flask, and the flask, stoppe

r, and liquid have a mass of 145.028 g. Find the density of gasoline in g/mL. Show work and report density with the appropriate number of significant figures.

1 answer:

Dominik [7]3 years ago

8 0

Answer:

0.690 g/mL

Explanation:

Density is the ratio of mass to volume:

gasoline mass = (flask+gas mass) - (flask mass) = 145.028 g -110.525 g

gasoline mass = 34.503 g

Density = (gas mass)/(gas volume) = (34.503 g)/(50.0 mL)

Density = 0.690 g/mL

_____

The least-precise number in this calculation is the sample volume, which has 3 significant figures. Thus, the answer is good only to 3 significant figures.

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To give 33.6 dm³ hydrogen gas at STP, 18.06 x 10²³ atoms of Na must react completely.

<h3>What is Mole concept ?</h3>

A mole is a unit of measurement used to measure the amount of any fundamental entity (atoms, molecules, ions) present in the substance.

As according to the given equation, 2 moles (ie 12.04 x 10²³ atoms) of Na-atoms produces 1 mole (22.4 ltr) of H₂-gas.

Hence, to produce 33.6 ltr (equivalent to 33.6 dm³) of H₂-gas produced by ;

= 12.04 x 10²³ atoms of Na / 22.4 ltr of H₂-gas x 33.6 ltr

= 18.06 x 10²³ atoms of Na

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2 years ago

Write a balanced equation for rusting. (Assume rust becomes Fe+2)

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The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.

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The electron dot structure for CI is

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The correct option is C.
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Answer: The de broglie wavelength is $2.037 \times 10^{-34} m$ .

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Calculate $\lambda = \frac{h}{p}$ as follows.

$\lambda = \frac{h}{p}$

where,

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p = momentum = $mass \times velocity$

Putting the values in the formula as follows.

$\lambda = \frac{h}{mass \times velocity}$

= $\frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}$

= $2.037 \times 10^{-34} m$

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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

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Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

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The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

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So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

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$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

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$\%C_E = \frac{E}{T} * 100\%$

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