The enthalpy of the creation of Methanol is negative. This means heat is released when the reaction proceeds. You would want to use a low temperature.
Answer:
See Explanation
Explanation:
According to the question, it is established that the pain caused by a lot of insect bites owes to the acidity of the insect bites.
If this is true, then there is an urgent need to neutralize the acidic insect bite using a basic substance.
Hence the pH of the creams used to treat insect bites should be basic(having a high pH).
2. The compound methene can not exist because it is supposed to be derived from the alkane called methane which contains only one carbon atom.
In order to form an alkene, there must be a double bond between two carbon atoms. This is not possible because there is only one carbon atom in the hypothetical methene. Hence, a compound named methene cannot exist.
Answer:
1.25 x 10^15Hz
Explanation:
c = frequency x wavelength
c is the speed of light, which is equal to 3.00 x 10^8 m / s
frequency = c /wavelength
= (3.00 x 10^8m /s) / (2.40 x 10^-5 cm x 1 m /100cm)
= (3.00 x 10^8 m/s) / 2.40 x 10^-7m
= 1.25 x 10^15/s 1 / s = 1Hz
So, the Frequency = 1.25 x 10^15Hz
I hope this helped :)
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.