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Xelga [282]
2 years ago
12

Chemical Reaction Type of reaction

Chemistry
1 answer:
solmaris [256]2 years ago
8 0

Answer:

Explanation:

2NaCl 2Na + Cl2         synthesis reaction

CaCl2 + F2 CaF2 + Cl2          oxidation-reduction

AgNO3 + KCl AgCl + KNO3      precipitation reaction

N2 + H2 2NH3      i dont know

C3H8 + 5O2 3CO2 +4 H2O Balanced Chemical Reactions and the Conservation of Mass

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At what phase of mitosis are two separate nuclei are formed as nuclear membranes form around
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Which of the following is a property of a pure substance?
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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.
alukav5142 [94]
1) Formulas:

a) mole fraction of component 1, X1

X1 =  number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles =  number of grams / molar mass


2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
7 0
2 years ago
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