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2 years ago

As the voltage across a resistor is increased, what happens to the resistance of the resistor?

Chemistry

2 answers:

vampirchik [111]2 years ago

7 0

<h3>HELLO THERE, I HOPE MY ANSWER WILL HELP YOU :)</h3>

<h2>ANSWER-</h2>

-We will be paying close attention to the Ohm’s Law here:-

>Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). >Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change.

iVinArrow [24]2 years ago

6 0

Answer:

your answer is wrong dear $\huge\boxed{\fcolorbox{cyan}{pink}{ello}}$

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If two metals both have the same color, have similar densities, and are about the same size, but one is shiny and the other is d

Papessa [141]

<span>If two metals both have the same color, have similar densities, and are about the same size, but one is shiny and the other is dull, they can be separated based on the lustrous physical property of a metal. Lustrous is the property of the metal to shine/gleam gently because of its particle arrangement of atoms. </span>

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2 years ago

A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr

Oksanka [162]

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = $t_{1/2}=25 minutes$

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{25 min}=0.02772 min^{-1}$

Initial concentration of A = $[A]_o=0.900 M$

Final concentration of A after 50 minutes = $[A]=?$

t = 50 minute

$[A]=[A]_o\times e^{-kt}$

$[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}$

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

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2 years ago

Hi friends!!!!!!!!!!!

Anna11 [10]

Answer:

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5 0

1 year ago

Read 2 more answers

What foods contain large amounts of sugars and starches?

ycow [4]

Carbs!! (or carbohydrates)

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2 years ago

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How many grams of lithium sulfide must be dissolved in 1600.0 g water to make a 2.0 molal solution?

klasskru [66]

Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal.
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g

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2 years ago