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MrRissso [65]
3 years ago
15

hat range, give the lower value and the higher value, respectively, for the following. (a) How many cubic meters of water are in

a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of 1000 kg/m3. How much mass does the water in the cloud have?
Chemistry
1 answer:
torisob [31]3 years ago
5 0

Answer:

a) 9424797000 m3

b) 9424797000000 liters

c) 9424797000000 kg

Explanation

Cylinder has a radius of 1 km= 1000 m.

Area Cylinder = π r^2  =  3,141599* (1000m)^2  = 3141599 m2

a) Volume Cylinder= area x Height=3141599 m2* 3000 m=  9424797000 m3

b) 1000 liters= 1m3,   9424797000 * 1000 = 9424797000000 liters

c) densiy = 1000 kg/m3, or 1 kg/ liter,   mass=  9424797000 lt * 1kg/lt =

9424797000 kg

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When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?
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Answer:

\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

q= mc\Delta T

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.

  • q= 47.1 J
  • m= 14.0 g
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Substitute these values into the formula.

47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)

Multiply the 2 numbers in parentheses on the right side of the equation.

47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c

47.1 \ J = (25.2 \ g*\textdegree C) *c

We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).

\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}

\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c

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The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

1.87 \ J/ g * \textdegree C =c

The heat capacity of the liquid is approximately 1.87 J/g°C.

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<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

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