Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

Answer;

Considering the types of bonds being created and severed in the transition state, the stability of this temporary structure is comparatively high.

Explanation;

• The reaction can be expressed as follows; NO(g)+F2(g)→NOF(g)+F(g)
• All chemical reactions, including exothermic ones, require activation energy to initiate. The activation energy is the least amount of energy needed for the reactants to come together, overcome opposing forces, and begin breaking bonds.
• When molecules encounter each other, their kinetic energy may be sufficient to stretch, bend, and eventually break bonds, resulting in chemical reactions.

Answer: The molecular formula will be

Explanation:

When percentages are provided, we assume the total mass to be 100 grams.

Thus, the mass of each element corresponds to the specified percentage.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1: convert given masses to moles.

Moles of C =

Moles of H =

Moles of O =

Step 2: For determining the mole ratio, divide each molar amount by the smallest number of moles calculated.

For C =

For H =

For O =

The resulting ratio of C: H: O= 4: 4: 1

Hence, the empirical formula obtained is

The empirical weight is calculated as = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now the molecular formula needs to be obtained.

The molecular formula can be derived as=

The true statement is B. With identical masses for both metals, the final temperature of the two will be more aligned with 498 K rather than 298 K, as iron's specific heat capacity is significantly greater than that of gold's.

Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2