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3 years ago

What is the current in a 160V circuit if the resistance is 5Ω?

Chemistry

2 answers:

oee [108]3 years ago

5 0

Answer:

use v=IR

Explanation:

so, 160=I*5

I = 160/5

=32 ampere

Wittaler [7]3 years ago

3 0

The current = 32 A

<h3>Further explanation</h3>

Electric current is a ratio between the voltage and the amount of resistance at the conductor.

Can be formulated

$\large{\boxed{\bold{I=\frac{V}{R}}}$

V = Voltage (volts)

R = resistance (ohms)

This formula is also known as Ohm's Law, which states that

the amount of current flowing is directly proportional to the voltage and inversely proportional to the resistance.

V=160 V

R=5Ω

$\tt I=\dfrac{160}{5}=32~A$

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What can you infer about the volume of a gas as absolute zero is approached?

schepotkina [342]

Answer:

The volume of a gas approaches zero as the temperature approaches absolute zero.

Step-by-step explanation:

You may have done a Charles' Law experiment in the lab, in which you measured the volumes of a gas at various temperatures.

You plotted them on a graph, and perhaps you were asked to extrapolate the graph to lower temperatures.

Your graph probably looked something like the one below.

There is clearly an x-intercept at some low temperature.

Inference: The volume of a gas approaches zero as the temperature approaches absolute zero.

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3 years ago

Solid potassium chlorate (kclo3) decomposes into potassium chloride and oxygen gas when heated. how many moles of oxygen form wh

IrinaVladis [17]

Answer:

0.665 mol of O₂.

Explanation:

1. Molecular chemical equation:

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

2. Mole ratios:

2 mol KClO₃ : 2 mol KCl : 3 mol O₂

3. Number of moles of KClO₃

Number of moles = mass in grams / molar mass

Molar mass of KClO₃ = 122.55 g/mol

Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol

3. Number of moles of O₂

As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.

3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃

x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂

Round to 3 significant figures: 0.665 mol of O₂ ← answer

3 0

3 years ago

Read 2 more answers

A scientist would like to make 450.0 mL of a 0.0200 M HCl by diluting a 7.00 M HCl solution. What volume of the 7.00 M HCl solut

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3 years ago

. In terms of potential energy, state why atoms tend to bond in nature?

Viefleur [7K]

Answer:

Explanation:

Atoms form chemical bonds to make their outer electron shells more stable. The type of chemical bond maximizes the stability of the atoms that form it. ... Covalent bonds form when sharing atoms results in the highest stability. Other types of bonds besides ionic and covalent chemical bonds exist, too.

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3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

Step 1: Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

Step 4: Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

Step 6: Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

Step 7: Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0

3 years ago