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creativ13 [48]
3 years ago
10

When you turn on the heat in the car, heat is being transferred by (1 point)

Chemistry
1 answer:
OLEGan [10]3 years ago
5 0
A. Radiation 

Air in a car is heated up with the radiator and then blown out with an air pump.
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How many atoms are in a sample of 72.8 g calcium (Ca)?
xxMikexx [17]
Atomic mass Ca = 40.078 a.m.u

40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??

72.8 x ( 6.02x10²³) / 40.078 =

4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms

hope this helps!
5 0
3 years ago
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Taya2010 [7]
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3 0
3 years ago
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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi
Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>

<u><em /></u>

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = <em>9.9652g of water</em>

<em />

I hope it helps!

4 0
3 years ago
What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

6 0
3 years ago
When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef
kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

\frac{Rc}{Rp} = \sqrt{\frac{Mp\\}{Mc} }

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

\frac{3.51}{1.98} = \sqrt{\frac{Mp\\}{44.01} }

1.77 = \frac{\sqrt{Mp} }{6.64}

11.75 = \sqrt{Mp}

Mp = 11.75^{2}

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0
3 years ago
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