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nevsk [136]
1 year ago
6

Acetylene is the substance burned in oxy-acetylene torches. Write a balanced equation for the complete oxidation reaction that o

ccurs when acetylene (C2H2) burns in air. Use the smallest possible integer coefficients.
Chemistry
1 answer:
nexus9112 [7]1 year ago
4 0

Answer: 2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products must be same as the mass of reactants.

For the conservation of mass, the number of atoms of each element must be same in reactants and products. Thus we need to balance the reaction by writing appropriate stochiometric coefficients.

All the hydrocarbons burn in oxygen to form carbon dioxide and water.Thus the complete balanced equation is:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

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What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?
dusya [7]
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

          6.86 = -㏒ Ka 

∴Ka = 1.38 x 10^-7

by using ICE table:

               H2PO4- →  H+  + HPO4
initial      0.4 m            0         0

change     -X                +X       +X

Equ       (0.4-X)               X          X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X)   by solving for X

∴X = 2.3x 10^-4 

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

        = -㏒ (2.3 x 10^-4)
 ∴PH  =  3.6

3 0
1 year ago
identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,
siniylev [52]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

3 0
2 years ago
Calculate the ratio of effusion rates of cl2 to f2 .
Lelechka [254]
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
4 0
2 years ago
A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

        \frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}

           V₂ = 124.91mL

6 0
2 years ago
Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord
Marina CMI [18]

Answer:

M_{acid}=0.563M

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the acid to obtain:

M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M

Regards!

5 0
2 years ago
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