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Jobisdone [24]
3 years ago
15

How much answer plz e fcprwg b4y[onbnbtmv';ck.

Chemistry
2 answers:
Ainat [17]3 years ago
4 0

Answer:

ANSWEERRRRR

Explanation:

aleksley [76]3 years ago
3 0
Oh my god man ask someone else
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Net Ionic equations
Goryan [66]
<span>Net Ionic equation: 3Zn^2+(aq) + 2PO4^3-(aq) ---> Zn3(PO4)2 (s)
hope it helps
</span>
4 0
4 years ago
If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus pro
frutty [35]

Answer:

81 molecules

Explanation:

The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;

C5H12 + 8O2 --> 5CO2 + 6H2O

The ratio of C5H12 to O2 from the above equation is 1 : 8.

Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;

3C5H12 + 24O2 --> 15CO2 + 18H2O

This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.

Total molecules is given as;

3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules

5 0
3 years ago
What is the percent by mass of nahco3 in a solution containing 10 g of nahco3 dissolved in 400 ml of h2o?
Marat540 [252]
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
7 0
3 years ago
Is the molecule SF4 polar or non polar?
krok68 [10]
S4 is polar molecule
6 0
4 years ago
What Group is Zn (zinc) in?<br> O A. 4<br> O B. 2A<br> O C. 12<br> O D. 30
Oxana [17]
I believe it’s (C.12)
3 0
4 years ago
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