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2 years ago

Maria walked on a beach made mostly of sand. She walked barefoot

Chemistry

1 answer:

OLga [1]2 years ago

4 0

A: The atoms in the sand were moving more quickly

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What substance is 52.4 % potassium and 47.6 % chlorine?

dimaraw [331]

Polymer sand is correct

4 0

2 years ago

Given 2.91 moles of a gas in a 500 milliliter-container, if the temperature is found to be 31 degrees Celsius, what is the press

kari74 [83]

Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K

P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm

3 0

2 years ago

Read 2 more answers

Atoms that have lost or gained one or more electrons are called ________.

SVETLANKA909090 [29]

Atoms that have lost or gained one or more electrons are called ions.

3 0

2 years ago

On average, Earth’s crust contains about 8.1% aluminum by mass. If a standard 12-ounce soft drink can contains approximately 15

MrMuchimi

Answer:

5400 cans

Explanation:

First we convert the total weight, 1 ton, to grams:

$1ton=1x10^{6}g$

Now we need to know the mass of aluminum:

$m_{Al}=\frac{10^{6}*8.1}{100} =81000g$

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

$n=\frac{81000g}{15g/can} =5400cans$

8 0

2 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

2 years ago