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Hatshy [7]
3 years ago
9

Why is a changing ocean temperature of only one or 2 degrees so concerning?

Chemistry
1 answer:
Elodia [21]3 years ago
4 0
Because even a little can drastically change sea life as well as continents and other islands temperatures.
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Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
ohaa [14]

<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>) = 0.021 M.
Ka(HCN) =  4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻] / [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] = x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M - x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>

4 0
3 years ago
What is the PH of a 1.3×(10)^-9 M HBr solution?
olga nikolaevna [1]

pH solution = 8.89

<h3>Further explanation</h3>

Given

The concentration of HBr solution = 1.3 x 10⁻⁹ M

Required

the pH

Solution

HBr = strong acid

General formula for strong acid :

[H⁺]= a . M

a = amount of H⁺

M = molarity of solution

HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1

Input the value :

[H⁺] = 1 x  1.3 x 10⁻⁹

[H⁺] = 1.3 x 10⁻⁹

pH = - log [H⁺]

pH = 9 - log 1.3

pH = 8.89

7 0
3 years ago
What is true concerning changes of state
Lynna [10]
I'm happy to help but do you have the options?
7 0
3 years ago
Identify this symbol.<br> Cl-
g100num [7]
Symbols on the periodic table represent an element. 

Cl is also known as chlorine. 
8 0
4 years ago
Determine the heat of reaction for the process TiO2(s) + 4HCl(g) TiCl4(l) + 2H2(g) + O2(g) using the information given below: Ti
Aleks [24]

Answer : The heat of reaction for the process is, 1374.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The main chemical reaction is,

TiO_2(s)+4HCl(g)\rightarrow TiCl_4(l)+2H_2(g)+O_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) Ti(s)+O_2(g)\rightarrow TiO_2(s)     \Delta H_1=-939.7kJ

(2) 2HCl(g)\rightarrow H_2(g)+Cl_2(g)    \Delta H_2=-184.6kJ

(3) Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)    \Delta H_3=-804.2kJ

We reversing reaction 1, 3 and multiplying reaction 2 by 2 and then adding all the equations, we get :

(1) TiO_2(s)\rightarrow Ti(s)+O_2(g)     \Delta H_1=939.7kJ

(2) 4HCl(g)\rightarrow 2H_2(g)+2Cl_2(g)    \Delta H_2=2\times (-184.6kJ)=-369.2kJ

(3) TiCl_4(l)\rightarrow Ti(s)+2Cl_2(g)    \Delta H_3=804.2kJ

The expression for heat of reaction for the process is:

\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(939.7kJ)+(-369.2kJ)+(804.2kJ)

\Delta H_{rxn}=1374.7kJ

Therefore, the heat of reaction for the process is, 1374.7 kJ

3 0
3 years ago
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