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fomenos
3 years ago
5

The sun shines with equal intensity on a farm field, an asphalt road, and the ocean. Which material will heat up most during the

day? Which will cool the slowest at night? Explain.
Chemistry
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

The sun shines with equal intensity on a farm field, an asphalt road and the ocean. So basically asphalt road are heated the most during the day

Explanation:

The heat of water is more than the specific heat of sand.

Therefore sand is hot.

.Water is reflecting solar radiation.  

The land retains more heat since the land absorb solar radiation.

Therefore the land surfaces warm faster.

Since water is a slow conductor of heat, it need more energy than the sand. so its temperature is increasing.  soil loses heat faster.  

The ocean heats slower than land , the land air is more warmer than lean air. After the sun set the land loses heat quickly .The air above it cools.

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In the drawing above, the hammer is acting as what type of simple machine? A. pulley B. screw C. wheel and axle D. lever
Mars2501 [29]
The drawing above is acting as a B) a screw.

4 0
2 years ago
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4 What volume of 1.45 M Ca(OH)2 is needed to react with 25.0 moles of H2SO4?
Shkiper50 [21]

Answer:

We need 17.2 L of Ca(OH)2

Explanation:

Step 1: Data given

Concentration of Ca(OH)2 = 1.45 M

Moles of H2SO4 = 25.0 moles

Step 2: The balanced equation

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4

Step 3: Calculate moles Ca(OH)2

For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4

For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4

Step 4: Calculate volume of Ca(OH)2

Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2

Volume Ca(OH)2 = 25.0 moles / 1.45 M

Volume Ca(OH)2 = 17.2 L

We need 17.2 L of Ca(OH)2

3 0
2 years ago
Use the unbalanced equation NH3 + O2 = NO + H2O
Elodia [21]
<span>The balance format is
4NH3+ 5O2 -------> 4NO + 6H2O </span>
7 0
2 years ago
The pressure on 150 mL of a gas is increased from 500 mm Hg to 700 mm Hg at constant temperature. What is the new volume of the
Ivanshal [37]

Answer:

Option A. 107 mL

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 150 mL

Initial pressure (P₁) = 500 mmHg

Final pressure (P₂) = 700 mmHg

Temperature = constant

Final volume (V₂) =?

The final volume of the gas can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

500 × 150 = 700 × V₂

75000 = 700 × V₂

Divide both side by 700

V₂ = 75000 / 700

V₂ = 107 mL

Therefore, the final volume of the gas is 107 mL.

4 0
3 years ago
What is the mass of 8.23 x 10^23 atoms of Ag
Gnom [1K]

Answer:

\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}

8.23 *10^{23}  *\frac{1 \ mol \ Ag}{6.022*10^{23} }

Condense into one fraction.

\frac{8.23 *10^{23}  }{6.022*10^{23} } \ mol \ Ag

1.366655596 \ mol \ Ag

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

  • Ag: 107.868 g/mol

Let's make another ratio using this information.

\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

Multiply by the number of moles we calculated.

1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

The moles of silver cancel out.

1.366655596 *\frac {107.868 \ g \ Ag}{1 }

1.366655596 * {107.868 \ g \ Ag}

147.4184058 \ g\ Ag

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

  • 147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

147 \ g\ Ag

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0
3 years ago
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