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Lelu [443]
3 years ago
8

the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

at is the final volume
Chemistry
1 answer:
Harman [31]3 years ago
4 0

Answer:

\boxed{\text{10 L}}

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}

Calculation:

\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}

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Write each element and elements of atoms of the compound Na2PO3F
Rudik [331]

The subscripts tell you <em>how many atoms</em> of an element are in one formula unit of a compound.

If there is no subscript, there is only one atom of the element.

In one formula unit of Na₂PO₃F, there are

Two atoms of sodium (Na)

One atom of phosphorus (P)

Three atoms of oxygen (O)

One atom of fluorine (F)

7 0
3 years ago
Which statement about cellulose is true? a) It’s a synthetic polymer. b) It’s a raw material used to make plastic. c) It’s produ
melisa1 [442]
<span>The correct answer is letter "d. It's used to make paper products." Cellulose is considered to be a major material used in creating a vast amount of paper products, such as papers, paperboards, and even cardboards. Cellulose can also be used as an important fiber as an important material in textiles.</span>
4 0
3 years ago
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A 12.82 g sample of a compound contains 4.09 g potassium (K), 3.71 g chlorine (Cl), and oxygen (O). Calculate the empirical form
Rina8888 [55]
The  empirical  formula of the  compound is  calculated  as  follows

first   calculate  the  mass  of  oxygen=  12-(4.09  +3.71)=  5.02g

then  calculate  the  moles  of  each  element,  moles  =  mass/  molar  mass

moles of   K  =  4.09g/39 g/mol(molar  mass  of K)  =  0.105  moles
moles  of Cl = 3.71g/35.5 g/mol(molar  mass  of Cl) =  0.105  moles
moles of  O  =  5.02g/ 16g/mol(molar  mass of  O) = 0.314  moles

then  calculate e  mole ratio by  dividing  each  mole  by  the  smallest  number  of  moles  (  0.105 moles)

K=0.105/0.105= 1
Cl=0.105 /0.105=1
O=  0.314/0.105=3

therefore  the  empirical   formula  = KClO3
7 0
2 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
3 years ago
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emmasim [6.3K]
<h3>Answer:</h3>

A. 860 kg

<h3>Explanation:</h3>

To answer the question we need to understand that;

  • Mass refers to the amount of matter in an object.
  • Weight, on the other hand, refers to the gravitational pull of an object to a given surface.
  • Mass is measured using a spring balance.

We also need to know that;

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  • Therefore, an object on the surface of the earth would have the same mass as on the surface of the moon.
  • In this case; the mass of the car remains the same on the outer space as on the back yard.
3 0
3 years ago
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