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Novay_Z [31]
3 years ago
5

In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 17

6 kg/h. Determine the percentage of excess air used during this process.
Chemistry
1 answer:
REY [17]3 years ago
4 0

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

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Which of the following elements will form negative ions? Check<br> all that apply.
AleksandrR [38]

Answer:

Explanation:

Elements on the right side of the periodic table are very likely to form negative ions -- all of those except elements in the 8th or 18th column (depending on how your periodic table is numbered).

K and Mg are on the left side, so they will not form negative ions.

They give up 1 (for K) electron and 2 (for Mg) electrons which will leave plus charges for the ions.

On the other hand S and I are on the right side of the periodic table. They will take on electrons and hence be charged with a minus.

5 0
3 years ago
Why does secondary succession usually occur more rapidly than primary succession.
trapecia [35]
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5 0
3 years ago
Please help.
Anna35 [415]

Answer:

9429.9

Explanation:

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4 0
3 years ago
11. Which is the solubility product expression for Na3PO4(s)?
dlinn [17]

Answer: B

Explanation: the 3 in Na3PO4(s) belong to the sodium atom (Na). so in any of these equations, the 3 would have to be with Na.

A- the 3 is along w the PO4, which would make it part of that bond

C- there is no 3 at all for Na in this choice, making it incorrect

D- Again, the 3 is placed on the other half of the bond

8 0
3 years ago
A 25.0-mL sample of an H2SO4 solution requires 50.0 mL of 0.150 M NaOH to completely react with the H2SO4. What was the concentr
Nitella [24]
V( H₂SO₄) = 25.0 mL in liters = 25.0 / 1000 = 0.025 L
 M(H₂SO₄) = ?

V(NaOH) = 50.0 mL = 50.0 / 1000 = 0.05 L 
M(NaOH) = 0.150 M

number of moles NaOH :

n = M x V

n = 0.150 x <span> 0.05 
</span>
n = 0.0075 moles of NaOH 

H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)

1 mole H₂SO₄ ---------- 2 mole NaOH
? mole H₂SO₄ ---------- 0.0075 moles NaOH

moles = 0.0075 * 1 / 2

= 0.00375 moles of H₂SO₄

M(H₂SO₄) = n / V

M = 0.00375 / <span> 0.025

</span>= 0<span>.15 M
</span>
hope this helps!

8 0
3 years ago
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