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3 years ago

A engineer measures the peak power output is 0.3227MW what is the peak power output in kilowatts ?

Chemistry

1 answer:

Lana71 [14]3 years ago

3 0

<h3>Answer:</h3>

322.7 kW

<h3>Explanation:</h3>

Power refers to the rate at which work is done.
Therefore; Power = Work done ÷ time
It is measured in joules per seconds or Watts

In this case, we are required to convert 0.3227 MW to kilowatts

We need to know that;

10^6 watts = 1 Megawatts(MW)
10^3 Watts = 1 kilowatts (kW)

Therefore;

10^3 kW = 1 MW

Therefore, the suitable conversion factor is 10^3kW/MW

Hence;

0.3227 MW is equivalent to;

= 0.3227 MW × 10^3kW/MW

= 322.7 kW

Thus, the peak power output is 322.7 kW

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Compute 9.3456+2140.56. Round the answer appropriately.

vichka [17]

This is just addition. Put 2140.56 on top, line up 9.3456 under it appropriately. Doing this will give you the answer: 2149.9056

5 0

3 years ago

I need help with the solution

Y_Kistochka [10]

Answer:

Explanation:

The empirical formula describes the simplest whole number ratio of each type of atom in a compound. To find this formula, you need to (1) convert grams of each element to moles (via their atomic masses) and then (2) find the ratio of each element (by dividing each molar value by the smallest mole value).

(Step 1)

Atomic Mass (Zn): 65.380 g/mol

Atomic Mass (C): 12.011 g/mol

Atomic Mass (O): 15.998 g/mol

10.40 grams Zn 1 mole
------------------------ x ------------------------ = 0.159 moles Zn
65.380 grams

1.92 grams C 1 mole
--------------------- x ----------------------- = 0.160 moles C
12.011 grams

7.68 grams O 1 mole
---------------------- x ------------------------ = 0.480 moles O
15.998 grams

(Step 2)

0.159 moles Zn / 0.159 = 1 atom Zn

0.160 moles C / 0.159 = 1 atom C

0.480 moles O / 0.159 = 3 atom O

The empirical formula is Zn₁C₁O₃. Therefore, the value that should be in the space is 1.

7 0

2 years ago

Help!!!

Mrac [35]

Answer:

Explanation:

Atomic mass is based on a relative scale and the mass of 12C

(carbon twelve) is defined as 12 amu; so, this is an exact number.

Why do we specify 12C? We do not simply state that the mass of a C atom is

12 AMU because elements exist as a variety of isotopes.

Carbon exists as two major isotopes, 12C, and 13C ( 14C exists and has

a half life of 5730 y, 10C and 11C also exist and their half lives are

19.45 min and 20.3 days respectively). Each carbon atom has the

same number of protons and electrons, 6. 12C has 6 neutrons, 13C has

7 neutrons, and 14C has 8 neutrons and so on. So, we must specify

which C atom defines the scale.

All the masses of the elements are determined relative to 12C.

Average Atomic Mass

Since many elements have a number of isotopes, chemists use average

atomic mass. On the periodic table the mass of carbon is reported as

12.011 amu. No single carbon atom has a mass of 12.011, but in a handful

of C atoms the average mass of a carbon atom is 12.011.

Why 12.011?

If a sample of carbon was placed in a mass spectrometer the

spectrometer would detect two different C atoms, 12C and 13C.

The natural abundance of 14C, 10C and 11C in geologic (i.e. old) samples is so low that we cannot

detect the effect these isotopes have on the average mass.

From the information collected from the mass spectrometer the average

mass of a carbon atom is calculated.

The mass of 12C is, of course, 12 amu.

13C is 1.0836129 times heavier than 12C; so, the mass of 13C is 13.003355

amu.

98.89% of the sample is 12C, and

1.11% of the sample is 13C

4 0

3 years ago

According to table g, which substance forms an unsaturated solution when 80. grams of the substance are stirred into 100. grams

e-lub [12.9K]

A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.

Unsaturated solution has solute in lower proportions than required to form a saturated solution.

Supersaturated solution has solute in amounts greater than a saturated solution.

We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.

The solubility of KI at 10 $^{0}C$ is 136 g/ 100 mL water

The solubility of $KNO_{3}$ at $10^{0}C$ is 21 g/100 mL water.

The solubility of $NaNO_{3}$ at $10^{0}C$ is 80 g/100 mL water.

The solubility of NaCl at $10^{0}C$ is 38 g/ 100 mL water.

So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at $10^{0}C$ .So, if we have 80g KI/ 100mL water it would be an unsaturated solution.

4 0

3 years ago

Read 2 more answers

A quantity of liquid methanol, CH3OH, is introduced intoa rigid 3.00-L vessel, the vessel is sealed, and the temperature israise

Tom [10]

Answer:

74,3 grams is the mass of methanol that was initially introduced into the vessel

Explanation:

CH3OH(g) ----> 1 CO(g) + 2 H2(g)

initial: Idk what i have (X) - -

react: initial - react react react

equilibrium: A concentration A conc. 0.426 M

in equilibrum in eq

See that in equilibrium are formed 2 moles of H2 with this concentration, 0,426M and you form 1 mol of CO, so the moles that are formed of H2 are the double of CO, because stoychiometry.

CH3OH(g) ----> 1 CO(g) + 2 H2(g)

initial: Idk what i have (X) - -

react: initial - react react react

equilibrium: A conc. in eq 0,213 M 0.426 M

So we have concentrations of products in equilibrium, and we have Kc, now we can find concentration of reactants in equilibrium

Kc = ([CO] . [H2]*2) / [CH3OH]

6,9X10*-2 = (0,213 . 0,426*2) / [CH3OH]

[CH3OH] = (0,213 . 0,426*2) / 6,9X10*-2

[CH3OH] = 0,560 M

You know that 1 mol which reacted has this concentration in equilibrium, 0,213 M so I can know what's my initial concentration of reactive

Initial - react = equilibrium

initial - 0,213M = 0,560M ---> Initial = 0,773 M

0,773M is initial concentration of CH3OH, but this is molarity, (moles in 1L). My volume is 3 L so

1 L _____ 0,773 moles

3 L _____ 3L . 0,773moles = 2,319 (the moles that i used)

Molar mass CH3OH = 32.06 g/m

Moles . molar mass = grams ---> 2,319 moles . 32,06 g/m = 74,3 g

8 0

3 years ago