NEED HELP AND WILL GIVE ALOT OF POINTS

What energy output objects work with the turbine?

MrMuchimi

The energy output object that works with the turbine is the alternator (generator)

8 0

3 years ago

5

Arada [10]

Answer:

They cause boys' and girls' bodies to develop

5 0

3 years ago

An oily liquid with a cheesy, waxy odor like that of goats or other barnyard animals is caproic acid, a compound containing carb

Zinaida [17]

The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O

We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:

How to determine the mass of C

Mass of CO₂ = 9.78 g
Molar mass of CO₂ = 44 g/mol
Molar of C = 12 g/mol
Mass of C =?

Mass of C = (12 / 44) × 9.78

Mass of C = 2.67 g

How to determine the mass of H

Mass of H₂O = 20.99 g
Molar mass of H₂O = 18 g/mol
Molar of H = 2 × 1 = 2 g/mol
Mass of H =?

Mass of H = (2 / 18) × 4

Mass of H = 0.44 g

How to determine the mass of O

Mass of compound = 4.30 g
Mass of C = 2.67 g
Mass of H = 0.44 g
Mass of O =?

Mass of O = (mass of compound) – (mass of C + mass of H)

Mass of O = 4.30 – (2.67 + 0.44)

Mass of O = 1.19 g

<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as follow:

C = 2.67 g
H = 0.44 g
O = 1.19 g
Empirical formula =?

Divide by their molar mass

C = 2.67 / 12 = 0.2225

H = 0.44 / 1 = 0.44

O = 1.19 / 16 = 0.074

Divide by the smallest

C = 0.2225 / 0.074 = 3

H = 0.44 / 0.0744 = 6

O = 0.074 / 0.074 = 1

Thus, the empirical formula of the compound is C₃H₆O

Learn more about empirical formula:

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6 0

1 year ago

In a nuclear fission reaction a heavy nucleus divides to form smaller nuclei and one or more neutrons. Many nuclei can undergo f

Effectus [21]

Answer:

$\frac{235}{92} U +$ $\frac{1}{0} n$ -------> $\frac{139}{36} Ba$ + $\frac{94}{56} Kr + 3\frac{1}{0} n$

$\frac{235}{92} U + \frac{1}{0} n -------> \frac{90}{38} Sr + \frac{143}{54} Xe + 3\frac{1}{0} n$

Explanation:

In equation 1, equating the mass number (A) on both sides.

A = 235 + 1 = A + 94 + 3*1

236 = A + 94 + 3

A = 236 - 94 = 3

A = 139

Equate the atomic numbers on both sides

92 + 0 = Z + 36 + 3*0

92 = Z + 36

Z = 92 - 36

Z = 56

In reaction 2, equating the mass number on both sides

235 + 1 = A + 143 + 3 *1

236 = A + 143 + 3

236 = Z + 146

Z = 90

Equatoing the atomic number of both sides

92 + 0 = Z + 54 + 3*0

92 = Z + 54

Z = 92 - 54

Z = 38.

7 0

3 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

3 years ago