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3 years ago

Which reacts with metals to form H2 gas

Chemistry

1 answer:

Ainat [17]3 years ago

5 0

Answer: acid react with metals to form H2.

Explanation: metals which are more electropositve than hydrogen will always displaced hydrogen from acids. The equation below show how hydrogen is displaced from acid

Zn + 2HCl -—> ZnCl2 + H2

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For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

3 years ago

How much pure water must mixed with 8 pints of 50 developer to produce a mixture that is 24%?

Galina-37 [17]

8:24 = 0.333(3) pints is one percent
0.333(3)* 100=33.333(3) pints will be 24% mixture with the water
33.3333-8=25.333 pints of water is required for producing 24% mixture

25.3333 pints of pure water and 8 pints of juce.

6 0

3 years ago

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th

mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene, Solvent 2 = water

$K_{p}$ = 2.7, $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL, weight of compound = 1 g

Extract = 3

Therefore, calculate the fraction remaining as follows.

$f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

= $[1 + 2.7(\frac{100}{10})]^{-3}$

= $(28)^{-3}$

= $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

= 1 - $4.55 \times 10^{-5}$

= 0.00001

or, = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$ .

7 0

3 years ago

Read 2 more answers

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

10. The statement that no two electrons in the same atom can have the

Fynjy0 [20]

Answer:

10.the pauli exclusive principle statues that, in an atom

or molecules, no two electrons can have same four electronic quantum numbers.as an orbit can contain a maximum of only two electrons.the two electrons must have opposing spins

3 0

3 years ago