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larisa [96]
3 years ago
6

Which of these is an element? A. carbon B. carbon dioxide C. air D. water

Chemistry
2 answers:
Sloan [31]3 years ago
8 0
I think it's B carbon dioxide
Kaylis [27]3 years ago
4 0
Elements are pure substances. Carbon is the only one that is pure.
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For which blocks of elements are outer electrons the same as valence electrons? For which are d electrons often included among v
Amanda [17]

Answer:

1. Group 1 — 3

2. Transition metals

Explanation:

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4 years ago
Someone help me please. It's due in 30 mins
max2010maxim [7]

Answer: permanent removal of hair by energy or heat

Explanation:

8 0
3 years ago
What is malic acid and phosphoric acid? ​
fomenos

Answer:

It is used to make medicine. People take malic acid by mouth for tiredness and fibromyalgia. In foods, malic acid is used as a flavoring agent to give food a tart taste. In manufacturing, malic acid is used to adjust the acidity of cosmetics.

Explanation:

hope it helps

5 0
3 years ago
Read 2 more answers
Titration Volume & Concentration
polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

0,0171L*\frac{0,161moles}{L} = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}} = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

5,5062x10^{-3}molHCl*\frac{1L}{0,304mol} = 0,0181L≡18,1mL

I hope it helps!

4 0
3 years ago
4 kmol Methane (CH4) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both
Svetach [21]

Answer:

Heat transfer = 3564 Jolues

The same value

Explanation:

The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):

Qtransfer = - mol x ΔHºc Qtransfer

So look up in appropiate reference table ΔHºc  and solve the problem:

ΔHºc  = - 891 kJ/mol

Qtransfer = - (4 x 10³ mol x  -891 kJ/mol ) = 3564 J

if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.

7 0
3 years ago
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