Charge can be transferred from one object to another in three way conduction, polarization, and friction
Whenever electrons are transferred between objects, neutral matter becomes charged and three ways this can occur are referred to as conduction, polarization, and friction conduction means when there is direct contact between materials that differ in their ability to give up or accept electrons when two different materials are rubbed together, there is a transfer of electrons from one material to the other material and this causes one object to become positively charged (the electron loser) and the other object to become negatively charged (the electron gainer)
Here given object is water + balloon; balloon + wall; tape then you rub a balloon against your clothes and it sticks to the wall, you are adding a surplus of electrons (negative charges) to the surface of the balloon the wall is now more positively charged than the balloon and we tested the conductivity of deionized water positively charged
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Copper wire is not an example of a pure element because although it's made by pure elements, it's not one itself. It's made by factories.
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Answer:
Molarity of solution is 1.10x10⁻³ M
Explanation:
Solute NaOCl
7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.
Molar mass of NaOCl = 74.45 g/m
Mol = Mass / Molar mass
7.4 g / 74.45 g/m = 0.099 moles
Density of solution = 1.12 g/mL
Density = Mass / volume
1.12g/mL = 100 g / volume
Volume = 100 g / 1.12g/mL = 89.3 mL
Molarity = mol /L
89.3 mL = 0.0893 L
0.099 moles / 0.0893 L = 1.10x10⁻³ M
Answer:
1. Hydrogen will diffuse faster.
2. The ratio of diffusion of hydrogen gas to that of the unknown gas is 4 : 1
Explanation:
Let the rate of diffusion of hydrogen gas, H2 be R1
Let the molar mass of H2 be M1
Let the rate of diffusion of the unknown gas be R2.
Let the molar mass of the unknown gas be M2.
Molar mass of H2 (M1) = 2x1 =2g/mol
Molar mass of unknown gas (M2) = 16 times that of H2
= 16 x 2 = 32g/mol
1. Determination of the gas that will diffuse faster. This is illustrated below:
R1/R2 = √(M2/M1)
R1/R2 = √(32/2)
R1/R2 = √16
R1/R2 = 4
Cross multiply
R1 = 4R2
From the above calculations, we can see that the rate of diffusion H2 (R1) is four times the rate of diffusion of the unknown gas (R2).
Therefore, hydrogen will diffuse faster.
2. Again, from the calculations made above, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is given by;
R1/R2 = 4
Therefore, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is:
4 : 1
