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3 years ago

How much heat in j is giving out when 85.0g of lead cools from 200.0 c to 10.0c

Chemistry

1 answer:

photoshop1234 [79]3 years ago

5 0

Answer:

q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.

Explanation:

The Specific Heat capacity of Lead is 0.128 $\frac{J}{g\ ^{0}C}$

This means, increase in temperature of 1 gm of lead by $1 ^{0}\ C$ will require 0.128 J of heat.

Formula Used :

$q = m.c.\Delta T$

q = amount of heat added / removed

m = mass of substance in grams = 85.0 g

c = specific heat of the substance = 0.128

$q = m.c.\Delta T$ = Change in temperature

= final temperature - Initial temperature

= 10 - 200

= - $190 ^{0}\ C$

put value in formula

q = - $85\times 0.128\times 190$

On calculation,

q = - 2067.2 J

- sign indicates that the heat is released in the process

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The permanent electric dipole moment of the water molecule 1H2O2 is 6.2 * 10-30 C m. What is the maximum possible torque on a wa

stellarik [79]

Answer:

3.1 x 10⁻²¹ Nm

Explanation:

When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:

torque (τ) = dipole moment vector (P) x electric field vector (E)

τ = P. E . sin θ

where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1

solve for τ

τ = 3.1 x 10⁻²¹ Nm

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3 years ago

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geniusboy [140]

Answer:

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3 years ago

A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place

kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8 cm³

mass of gold cube = 20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

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3 years ago

What is the pH of a 6.7 x 10^−5 M H+ solution?

lyudmila [28]

Answer:

pH = 4.17

Explanation:

According to the molar concentration you stated, pH of the solution is: 4.17

Remember that pH = - log [H⁺]

and [H⁺] = 10^-pH

When:

pH > 7 → Basic solution

pH = 7 → Neutral solution

pH < 7 → Acid solution

3 0

3 years ago

Read 2 more answers

The concentration of carbon monoxide (CO), a common air pollutant, is found in a room to be 5.7 x 10^-3 mg/cm^3. How many grams

damaskus [11]

The amount, in mg, of CO present in the room will be 191,520 mg.

<h3>Stoichiometric problem</h3>

The concentration of the gas in the room is 5.7 x $10^{-3}$ mg/cm3.

The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.

We can obtain the volume of the room as:

350 x 300 x 320 = 33,600,000 cm3

The concentration is in mg/cm3, meaning that it is mass/volume.

Thus:

mass = concentration x volume = 5.7 x $10^{-3}$ mg/cm3 x 33,600,000 cm3

= 191,520 mg

The mass of CO in the room is 191,520 mg

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

5 0

1 year ago