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2 years ago

7

Static charges can be applied to neutral objects by friction, induction or conduction. What do all of these methods utilize to c

reate this charge?
a.protons
b.electrons
c.heat energy

1 answer:

MatroZZZ [7]2 years ago

5 0

Electrons are valence and free moving so they take place in charge transfer

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How many of moles in 162 grams of H2SO4

blondinia [14]

Answer:

0.010195916576195

Explanation:

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2 years ago

The element hydrogen adom is made up of wich kinds of adams

egoroff_w [7]

the element hydrogen atom is made up of oxygen atoms.

3 0

2 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

Helpppppp helpppppp helppppp

soldier1979 [14.2K]

where is the diagram?

without the diagram i can't help

7 0

2 years ago

1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.

Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of $CH_3OH$ = 26.5 g

Molar mass of $CH_3OH$ = 32.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{26.5\ g}{32.04\ g/mol}$

$Moles\ of\ CH_3OH= 0.8271\ moles$

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

$m=\frac {0.8271\ moles}{0.735\ kg}$

<u>Molality = 1.13 m</u>

4 0

2 years ago