<h2>
An acidic solution has > </h2>
Explanation:
pH is the negative logarithm of hydronium ion concentration present in a solution.
If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9. The solution has more than
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14. The solution has more than
The solution having pH equal to 7 is termed as neutral solution.The solution has equal concentration of and
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Answer:
B. exothermic; leaving
Explanation:
The exothermic process releases heat, which causes the surrounding area to increase in temperature.
Your hand is releasing heat and makes the temperature of the ice cube increase, to where it melts.
The balanced chemical equation is this:
Cl2(aq) + 2Br-(aq) --> 2Cl- + Br2(aq)
Looking at each substance involved in the reaction:
Cl2 has 0 charge and gains a charge of -1 after the reaction
Br- has a charge of -1 but turns to 0 after the reaction
Cl2 gained an electron which means it is the reduced substance
Br- lost an electron which means it is the oxidized substanced
Consequently, Cl2 is the oxidizing agent and Br- is the reducing agent.
Answer:
d. 103.3
Explanation:
In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:
In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.
Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.
The energy of the 434 nm emission line is 4.58×10¯¹⁹ J (Option A)
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Wavelength (λ) = 434 nm = 434×10¯⁹ m
- Planck's constant (h) = 6.626×10¯³⁴ Js
- Speed of light (v) = 3×10⁸ m/s
- Energy (E) =?
<h3>How to determine the energy </h3>
The energy of the 434 nm emission line can be obtained as follow:
E = hv / λ
E = (6.626×10¯³⁴ × 3×10⁸) / 434×10¯⁹ 9.58×10¹⁴
E = 4.58×10¯¹⁹ J
Thus, the energy of the 434 nm emission line is 4.58×10¯¹⁹ J
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