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pychu [463]
3 years ago
10

- How much power does it take to lift a 1,000 N load 10 m in 20 s?

Chemistry
1 answer:
Mariana [72]3 years ago
5 0

Answer:

"500 Joule/sec" is the right answer.

Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

Now,

The power will be:

=  \frac{Force\times Velocity}{Time}

On putting the values, we get

=  \frac{1000\times 10}{20}

=  \frac{10000}{20}

=  500 \ Joule/sec

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Law of electrolysis was given by<br>a.lamark. b.Ostwald<br>c.Faraday. d.Arrhenius<br>​
Brrunno [24]

Law of electrolysis was given by Faraday.

He discovered that,the amount of chemical change or decomposition is exactly proportional to the quantity of electricity that passes in solution,

4 0
3 years ago
How do you write the balanced equation ​
katrin [286]

Answer:

See BELOW!

Explanation:

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6 0
3 years ago
16. A sample of nitrogen gas,
Lana71 [14]

find mol of N2 present using gas law equation

PV = nRT

P = pressure = 688/760 = 0.905 atm.

V = 100mL = 0.1L

n = ???

R = 0.082057

T = 565+273 = 838

Substitute:

0.905*0.1 = n*0.082057*838

n = 0.0905 / 68.76

n = 0.00132 mol N2

Molar mass N2 = 28 g/mol

0.00132 mol = 0.00132*28 = 0.037g N2 gas

4 0
3 years ago
Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo
Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

7 0
3 years ago
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