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Dvinal [7]
2 years ago
6

If the hydrogen ion concentration, (H+), in a solution is 5.39 x 10-8 M,

Chemistry
1 answer:
FrozenT [24]2 years ago
6 0

Answer: 1.86*10^-7M

Explanation:

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Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?
mylen [45]

Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is  the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

Therefore, Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

4 0
2 years ago
True or false: Fungi are able to carry out photosynthesis, similar to plants
QveST [7]

Answer:

false

Explanation:

7 0
2 years ago
Read 2 more answers
An organic compound, which has the empirical formula C12H25 has an approximate molar mass of 338 g/mol. What is its probable mol
Marta_Voda [28]

Answer:

C24H50

Explanation:

The empirical fomula's molar mass is 169.25 g/mol.

We know the molecular formula's molar mass is 338 g/mol.

338/169.25= 1.99 or approximately 2

8 0
3 years ago
What is the chemical formula of tin(IV) chloride pentahydrate?
Vladimir [108]

Answer:

SnCl4 * 5H2O

Explanation:

4 0
3 years ago
A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera
Grace [21]

Answer: The initial temperature of the iron was 515^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of iron = 360 g

m_2 = mass of water = 750 g

T_{final} = final temperature = 46.7^0C

T_1 = temperature of iron = ?

T_2 = temperature of water = 22.5^oC

c_1 = specific heat of iron = 0.450J/g^0C

c_2 = specific heat of water= 4.184J/g^0C

Now put all the given values in equation (1), we get

-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]

T_i=515^0C

Therefore, the initial temperature of the iron was 515^0C

4 0
3 years ago
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