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klio [65]
3 years ago
9

A graduated cylinder( approximate as a regular cylinder) has a radius of 1. 045 cm and a high of 30.48 cm. what is the volume of

the cylinder centimeter 3( use 3.141 for ) round to the nearest tenths
Chemistry
1 answer:
4vir4ik [10]3 years ago
5 0

Given parameters:

Radius of cylinder  = 1.045cm

Height of cylinder  = 30.48cm

Unknown:

Volume of the cylinder  = ?

A cylinder is a solid body. The volume is usually calculated using the mathematical expression below;

     Volume of cylinder  = π r² h

  π  = 3.141

   r  = radius of cylinder

   h = height of cylinder

Now, input the parameters and solve;

          Volume of cylinder  = 3.141 x 1.045² x 30.48

                                            = 104.547cm³

To the nearest tenth gives 104.5cm³

The volume of the cylinder is 104.5cm³

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Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}

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[Mg^2+] = \dfrac{0.003474 \ M\times (26.23  - 20.49 )mL}{10.0 \ mL}

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Mass of Ca²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

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= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

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3 years ago
a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses
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Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

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P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa

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When barium reacts with tellurium to form an ionic compound, each metal atom loses electron(s) and each nonmetal atom gains elec
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<u>Answer:</u> The ionic compound formed is baF_2 (barium fluoride)

<u>Explanation:</u>

Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.

Taking the metal as barium and non-metal as fluorine.

Barium is the 56th element of the periodic table having electronic configuration of [Xe]6s^2

This element will loose 2 electrons and will form Ba^{2+} ion

Fluorine is the 9th element of the periodic table having electronic configuration of [He]2s^22p^5

This element will gain 1 electron and will form F^{-} ion

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

Hence, the ionic compound formed is baF_2 (barium fluoride)

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