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3 years ago

Convert 15.78 g of H2O2 to mols. Convert 15.3 mols Au to g

Chemistry

1 answer:

iren2701 [21]3 years ago

8 0

Answer:

.46 moles H2O2

3,014 grams Au

Explanation:

H2O2:

15.78g (1 mol/34g) = .46 moles H2O2

Au:

15.3mol (197g/mol) = 3,014 grams Au

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7. Explain how magnetic potential energy can be transformed into kinetic energy. (1 point)

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The potential energy by the magnetic field can turn into kinetic energy once the field is moving from the S pole to the N pole when it reaches the N pole it is potential energy when it exits the S pole it is kinetic energy.

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2 years ago

If a car passes a pedestrian, a change in pitch is __________.

enot [183]

Given two questions:

1) If a car passes a pedestrian, a change in pitch is ______________.
The answer is the change in pitch is perceived by the pedestrian since he is the one in a relatively constant position compared to the car passing.

2) In the Doppler Effect lab, which statement best describes what you demonstrated about speed and pitch?

The answer is 'speed and direction affect pitch'.

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3 years ago

Read 2 more answers

Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg

FrozenT [24]

Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

8 0

3 years ago

At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2

saul85 [17]

2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
K = 0.0556

4 0

3 years ago

Anybody good at chemistry?

Luba_88 [7]

Answer:

Explanation:

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C - 40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C × 55°C

Q = 42.35 j

Given data:

Mass = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C - 20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C × 80°C

Q = 95.48 j

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C - 60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C × -30°C

120 j = m × -26.91 j/g

m = 120 j / -26.91 j/g

m = 4.46 g

negative sign show heat is released.

Given data:

Mass of ice = 1.5 g

Change in temperature = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C × ΔT

30.0 j = 0.753 j/°C × ΔT

30.0 j /0.753 j/°C = ΔT

39.84 °C = ΔT

3 0

3 years ago