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Misha Larkins [42]
3 years ago
9

how much current would be measured in a circuit if the light bulb has a resistance of 6 ohms and a voltage of 36 volts

Chemistry
1 answer:
Stels [109]3 years ago
8 0

Answer:

The right response is "6 A". A further explanation is given below.

Explanation:

The given values are:

Resistance,

R = 6 ohms

Voltage,

V = 36 volts

As we know,

⇒  V=IR

then,

⇒  I=\frac{V}{R}

On substituting the values, we get

⇒     =\frac{36}{6}

⇒     =6 \ A

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Explanation:

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3. reduce the scattering by two orders.

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Can someone explain how you get the answers to these questions??<br> i really need help !!!
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<h3>What are balanced chemical equations?</h3>

Balanced chemical equations in which the moles of atoms of elements taking part in a reaction are balanced on both sides of the equation.

The balancing of chemical equations follows the law of conservation of mass which states that matter can neither be created nor destroyed.

In balancing of chemical equations, the following steps are follows:

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In conclusion, a balanced chemical equation obeys the law of conservation of mass.

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5 0
2 years ago
Convert this temperature from F to C<br><br> 26.6°c<br> 93.600<br> 62.2°C<br> 5.7°C
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3 0
3 years ago
Na3po4 dissolves in water to produce an electrolytic solution. what is the osmolarity of a 2. 0 × 10-3 m na3po4 solution?
vichka [17]

When Na3po4 dissolves in water to produce an electrolytic solution. The osmolarity of a 2. 0 × 10-3 m Na3po4 solution is 0.008osmol/L.

Osmolarity is defined as the number of osmoles of solute particles per unit volume of the solution.

In other words osmolarity is the multiple if molarity

Osmolarity = i× molarity

Here i represents the van't Hoff factor,

Na_{3}PO_{4} ⇒ 3Na^{+} + PO_{4} ^{-}

3  Moles of Na_{} ^{+} + 1 mole PO_{4} ^{-} = 4

The number of moles of particles of solute produced in solution are actually called osmoles.

As a result, the van't Hoff factor will be equal to

i=4 Moles ions produced (osmoles) 1mole Na_{3} PO_{4} .dissolved =4

Since we know that,

Na_{3} PO_{4} = 2.0 * 10^{-3} M

Osmolarity =

4*2.0*10^{-3} M = 8.0 * 10^{-3}  osmol L -10s

Thus, the Osmolarity of given solution is 0.008 osmol/L.

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2 years ago
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Answer:....................................

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