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Murljashka [212]
3 years ago
11

An iguana runs back and forth along the ground. The horizontal position of the iguana in meters over time is shown

Physics
1 answer:
pochemuha3 years ago
3 0

Answer:

MY

7-

Col

6-

SA

5+

4-

MY

Pro

3

Pro

2

Tea

+

→ t(s)

1

2

3

4

5

6

7

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Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
4 years ago
What mass of silver (in grams) is solidified when 749 joules of heat are released by a sample of molten silver at its freezing p
Anastasy [175]

heat released Q = 749 joules

heat of fusion of silver L = 109 J/g

Here phase of silver is changing from liquid to solid

so temperature will remain same

all heat will be released due to its phase change

and in this case we use Q=mL

where m is the mass of silver in gram

Q= mL

749 = m * 109

m = 749/109

m = 6.87 gram

4 0
4 years ago
Read 2 more answers
Estimate the total number of bacteria and other prokaryotes in the biosphere of the earth. (assume the bacteria are found to a d
Fofino [41]
Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:

V = 1/3*πh²(3r - h)
Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³

Thus, the total amount of bacteria is:

2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria
7 0
4 years ago
A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
Airida [17]

Answer:

  • The work made by the gas is 7475.69 joules
  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

E= c_v n R T

where c_v its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

\Delta E = \Delta Q - \Delta W

where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

\Delta Q = \Delta W

7 0
3 years ago
What is the equation for the enthalpy of vaporization?
Natalka [10]

Answer: Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.

4 0
3 years ago
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