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nataly862011 [7]
3 years ago
12

replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs can save a lot of energy. Calcul

ate the amount if energy saved over 10 h when one 60 W incandescent lightbulb is replaced with an equivalent 18 W compact fluorescent lightbulb
Chemistry
1 answer:
Fantom [35]3 years ago
8 0
The difference between the wattage is 60-18=42W. The saving in energy over 10 hrs is 10×42=420 watt-hours=0.42kWh
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What do we mean when we say atoms bond together?
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When two or more atoms chemically bond together, they form a molecule. Sometimes the atoms are all from the same element.

A Compound.

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For example, when three oxygen atoms bond together, they form a molecule of ozone (O3). If a molecule forms from atoms of two or more different elements, we call it a compound.

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Calculate the number of molecules present in 3.50 mol feo2. 2.11 x 1024 molecules 2.107 x 1023 molecules 3.50 x 1023 molecules
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Calculate the number of molecules present in 3.50 mol feo2. 2.11 x 1024 molecules 2.107 x 1023 molecules 3.50 x 1023 molecules <u>3.5*(6.02*10^23)=2.107*10^24 molecules</u>

<h3>What is molecules?</h3>

A molecule is the smallest unit of a substance that keeps its content and properties. It is made up of two or more atoms that are joined together by chemical bonds. Chemistry is built on molecules. The element symbol and a subscript indicating the number of atoms are used to identify molecules.

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6 0
2 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
3 years ago
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