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hjlf
3 years ago
12

Carbon dioxide is a non-polar molecule true or false​

Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

True

Explanation:

Due to the arrangement of the molecule, a carbon dioxide molecule is non-polar.

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#1: Jim, Jane, Ann, and Bill measure an object’s length, density, mass, and volume, respectively. Which student’s measurement mi
STatiana [176]
1. The <span>student with a measurement that might be in centimeters is A. Bill.
2. C</span><span>entimeters  in 0.05 kilometers is </span>C. 5,000
solution:
0.05 km x (1000 meters/ 1 km)
= 50 meters x (100 cm/ 1 meter) 
=5000 cm
7 0
3 years ago
Read 2 more answers
A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70
kakasveta [241]

Explanation:

The given data is as follows.

  P_{atm} = 98.70 kPa = 98700 Pa,  

      T = 30^{o}C = (30 + 273) K = 303 K

      height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = 13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}

                = 13534 kg/m^{3}

The relation between pressure and atmospheric pressure is as follows.

             P = P_{atm} + \rho gh

Putting the given values into the above formula as follows.

            P = P_{atm} + \rho gh

               = 98700 Pa + 13534 \times 9.81 \times 0.03 m

               = 102683.05 Pa

               = 102.68 kPa

thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.

8 0
4 years ago
hurry please! avogadro's law relates the volume of a gas to the number of moles of gas when temperature and pressure are constan
DIA [1.3K]

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:

V1/n1 = V2/n2

V/2 = 2V/n2

Cross multiply

V x n2 = 2 x 2V

Divide both side by V

n2 = (2 x 2V)/V

n2 = 2 x 2

n2 = 4 moles

Therefore, 4 moles of the gaseous product were produced.

5 0
3 years ago
As Period of Revolution (Increases or Decrease),Your age (Increases or Decreases or Stays the same)
xxMikexx [17]

Answer:

increase , my age increase

Explanation:

hope it helps..

6 0
3 years ago
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
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