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4 years ago

what natural resources are used to make a computer? what natural resource am i using to operate a computer?

1 answer:

7 0

Oil, coal, natural gas, metals, stone and sand are natural resources. Other natural resources are air, sunlight, soil and water. ... Natural resources also are the raw materials for making products that we use everyday from our toothbrush and lunch box to our clothes, cars, televisions, computers and refrigerators.

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"what is the concentration of positive charge and mass in rutherford's atomic model called?"

Tamiku [17]

The concentration of positive charge and mass in Rutherford's atomic model is called the nucleus. Rutherford's experiments involving the use of alpha particle beams directed onto thin metal foils demonstrated the existence of the nucleus. The nucleus of an atom contains positively charge particles called protons and other uncharged particles called neutrons. According to this model most volume of an atom is made up of an empty space.

3 0

3 years ago

What is the independent, dependent, and controlled variable?

attashe74 [19]

Independent Variable: amount of sunlight given

Dependent Variable: How tall the plants grow

Control: The plant given no sunlight

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4 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

You suspect one of the products of a single replacement is hydrogen gas. How might you experimentally confirm your prediction? C

Nataly_w [17]

The best option i think is b would be my best answer

6 0

3 years ago

How many moles of Carbon are in 3.06 g of Carbon

natta225 [31]

Answer:

$\boxed {\boxed {\sf 0.255 \ mol \ C }}$

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

$\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Since we are converting 3.06 grams to moles, we multiply by that value.

$3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

$3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}$

$3.06 * \frac{1 \ mol \ C}{12.011 }$

$\frac {3.06}{12.011 } \ mol \ C$

$0.25476646 \ mol \ C$

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

$0.255 \ mol \ C$

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0

3 years ago