Answer: Option (d) is the correct answer.
Explanation:
The given compound is [tex]N_{2}O_{3}[/tex]. It contains two nitrogen atoms and three oxygen atoms.
Therefore, according to naming nomenclature "di" will be added before the name of nitrogen. Whereas a prefix "tri" will be added before the name of oxygen atom.
Hence, name of the compound [tex]N_{2}O_{3}[/tex] is dinitrogen trioxide.
The correct name for N2O3 is dinitrogen trioxide. This is demonstrated by the di- prefix in dinitrogen indicating two nitrogen atoms, and the tri- prefix in trioxide indicating three oxygen atoms. The correct name for N2O3 is dinitrogen trioxide. Nitric oxide (NO) and nitrous oxide (N2O) are different compounds.
Explanation:The correct name for the chemical compound with the formula N2O3 is dinitrogen trioxide.
this can be broken down as follows: the prefix 'di-' in dinitrogen indicates that there are two nitrogen atoms, and the prefix 'tri-' in trioxide indicates there are three oxygen atoms. Therefore, the formula correctly matches the name since there are two Nitrogen atoms and three Oxygen atoms.
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A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution? 1.0 × 10–5 M OH– 1.0 × 10–14 M OH– 1.0 × 105 M OH– 1.0 × 10–9 M OH–
Answer:
1.0 x 10⁻⁹ M OH⁻.
Explanation:
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 1.0 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(1.0 x 10⁻⁵ M) = 1.0 x 10⁻⁹ M.
So, the right choice is: 1.0 x 10⁻⁹ M OH⁻.
Answer:
1.0 x 10-9M OH-
Explanation:
I just took the test and this was correct
Calculate the kinetic energy in j of an electron moving at 6.00 × 106 m/s. The mass of an electron is 9.11 × 10-28 g.
Answer:
The kinetic energy in J of an electron moving at 6.00 × 10⁻⁶ m/s is:1.64 × 10 ⁻³⁸ J
Explanation:
1) Data:
a) KE =?
b) v = 6.00 × 10⁻⁶ m/s.
c) m = 9.11 × 10⁻²⁸ g.
2) Formula:
KE = (1/2) mv²3) Solution:
KE = (1/2) = (1/2) × 9.11 × 10⁻²⁸ g × ( 6.00 × 10⁻⁶ m/s)² = 163.98 × 10 ⁻⁴⁰ J KE = 1.64 × 10 ⁻³⁸ JAnswer: The kinetic energy of the electron is [tex]1.64\times 10^{-17}J[/tex]
Explanation:
To calculate the kinetic energy of the electron, we use the equation:
[tex]E=\frac{1}{2}mv^2[/tex]
where,
m = mass of the electron = [tex]9.11\times 10^{-28}g=9.11\times 10^{-31}kg[/tex] (Conversion factor: 1 kg = 1000 g)
v = speed of the electron = [tex]6.00\times 10^6m/s[/tex]
Putting values in above equation, we get:
[tex]E=\frac{1}{2}\times 9.11\times 10^{-31}kg\times (6.00\times 10^6m/s)^2\\\\E=1.64\times 10^{-17}J[/tex]
Hence, the kinetic energy of the electron is [tex]1.64\times 10^{-17}J[/tex]
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacity is 0.505KJ/°C, what will be the delta H for the solution process.
25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)
ExplanationThe process:
[tex]\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)[/tex].
How many moles of this process?
Relative atomic mass from a modern periodic table:
K: 39.098;N: 14.007;O: 15.999.Molar mass of [tex]\text{KNO}_3[/tex]:
[tex]M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}[/tex].
Number of moles of the process = Number of moles of [tex]\text{KNO}_3[/tex] dissolved:
[tex]\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}[/tex].
What's the enthalpy change of this process?
[tex]Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}[/tex] for [tex]0.212162\;\text{mol}[/tex]. By convention, the enthalpy change [tex]\Delta H[/tex] measures the energy change for each mole of a process.
[tex]\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}[/tex].
The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.
To calculate the change in enthalpy for the solution process of KNO3, first find the heat transferred, q, through the formula q = m * C * ΔT. Then, calculate ΔH by dividing the heat transferred by the quantity of substance (in moles) involved, obtained from the given mass and the molar mass of KNO3.
Explanation:To calculate the change in enthalpy (ΔH) in the solution process of KNO3, you first need to figure out the amount of heat transferred during the process. This is done by using the formula: q = m * C * ΔT. In this case, the heat lost to the surrounding (q) equals the mass (m), which here is 21.45 g, times the specific heat capacity (C), which is given as 0.505 KJ/°C, times the change in temperature (ΔT), which is 25.00°C - 14.14°C.
After calculating q, ΔH can be calculated by taking into account the quantity of substance involved, which is the molar mass of KNO3. ΔH is reported in KJ/mol, so to get to ΔH, you'd find the molar mass of KNO3 (101.1 g/mol), figure out how many moles 21.45g represents, and then report the heat per mole. Therefore, the actual ΔH would depend on the specific values you use in these calculations.
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Find the Ka of nitrous acid given that a 0.20 M solution of the acid has a hydrogen ion concentration of 2.8*10-3 M. HNO2 (aq) ? H+ (aq) + NO2- (aq)
Answer:
3.92 x 10⁻⁵.
Explanation:
∵ [H⁺] = √(Ka.c)
∴ Ka = [H⁺]²/c = (2.8 x 10⁻³)²/(0.20) = 3.92 x 10⁻⁵.
What is the rate law for step 1 of this reaction? express your answer in standard masteringchemistry notation. for example, if the rate law is k[a][c]3 type k*[a]*[c]^3?
Answer:
rate = k*[A]^3
Explanation:
The following information is missing in the question:
Consider the following elementary steps that make up the mechanism of a certain reaction
1. 3A -> B+C
2. B+2D -> C+F
In elementary reactions, the order of each reactant in the rate law is equal to the coefficient in the balanced equation. Therefore, for the first step:
rate = k*[A]^3
The rate law for step 1 of the reaction is rate = k[CH3CH₂Cl]. The rate law is consistent with the experimentally derived rate law for the overall reaction. The rate constant (k) is 1.6 × 10^-6 s^-1.
Explanation:The rate law for step 1 of this reaction can be determined by comparing experiments and observing how changing the concentration of reactants affects the rate. Comparing experiments 2 and 3 shows that doubling the concentration of [CH3CH₂Cl] doubles the reaction rate. Similarly, comparing experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate. This behavior indicates that the reaction rate is directly proportional to [CH3CH₂Cl].
Based on this information, the rate law for step 1 is rate = k[CH3CH₂Cl]. Thus, the rate law for step 1 is consistent with the experimentally derived rate law for the overall reaction, which is rate = k[CH3CH₂Cl].
The rate constant (k) can be calculated using any row in the table. Selecting Experiment 1, we can set up an equation to solve for k:
1.60 × 10^-8 M/s = k(0.010 M)
k = 1.6 × 10^-6 s^-1
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