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2 years ago

6

Increase the brightness of the light for the colors red, yellow, orange, and green. What effect did the brightness of the light

have on the metal?
(science)

2 answers:

vichka [17]2 years ago

7 0

Answer:

umm i really do not know sorry

NNADVOKAT [17]2 years ago

5 0

Hmmm.......... .......

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For the following reaction, calculate how many moles of each product are formed when 4.05 g of water is used.

timurjin [86]

Answer:

A. 0.225 mole of H₂

B. 0.113 mole of O₂.

Explanation:

We'll begin by calculating the number of mole in 4.05 g of water (H₂O). This can be obtained as follow:

Mass of H₂O = 4.05 g

Molar mass of H₂O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mole of H₂O =?

Mole = mass /Molar mass

Mole of H₂O = 4.05 / 18

Mole of H₂O = 0.225 mole

Next, the balanced equation.

2H₂O —> 2H₂ + O₂

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂ and 1 mole of O₂.

A. Determination of the number of mole hydrogen produced.

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂.

Therefore, 0.225 mole of H₂O will also produce 0.225 mole of H₂.

B. Determination of the number of mole oxygen produced.

From the balanced equation above,

2 moles of H₂O produced 1 mole of O₂.

Therefore, 0.225 mole of H₂O will produce = (0.225 × 1)/2 = 0.113 mole of O₂.

Thus, 0.113 mole of O₂ is produced.

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3 years ago

Lisa [10]

Answer:

Hi! I'm feeling great and happy! Life's been good! How are you? :) Hope life's good for you and that you are staying safe.

Explanation:

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3 years ago

9.81 m/s^2 to ft/s^2

seropon [69]

9.81m* 100cm/1m*1in/2.54 cm* 1ft/12in=32.19 ft/s^2

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3 years ago

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What is the average atomic weight of silver?

Natali5045456 [20]

Answer:

107.8682 u would be the answer

Explanation:

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3 years ago

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A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago