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Bess [88]
3 years ago
6

Resistance of a material being scratched in known as

Chemistry
1 answer:
Nastasia [14]3 years ago
5 0
Friction is the answer

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How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
How many Earths will we need to have enough resources if the whole world lived like Americans?
Oxana [17]

Answer:

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Explanation:

because yeah

4 0
3 years ago
Read 2 more answers
Please help!!!!!<br> Please help!!!!!
lesantik [10]

Answer:

1.B

2.A

3. B

Explanation:

1. A chemical bond is the physical phenomenon of chemical substances being held together by attraction of atoms to each other through sharing

2. so 2 is electrons of one atom are transferred permanently to another atom.meaning the answer is A

3. is When atoms combine by forming covalent bonds, the resulting collection of atoms is called a molecule. We can therefore say that a molecule is the simplest unit of a covalent compound.

3 0
3 years ago
Read 2 more answers
Is matter lost when a candle is burned
zlopas [31]
No it is redistributed and the state changes to gas and liquid
5 0
3 years ago
Read 2 more answers
A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for
zloy xaker [14]

Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = \frac{72.15}{100}

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = \frac{27.85}{100}

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]

A=85.47

Therefore, the atomic weight of the metal would be 85.47.

6 0
3 years ago
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