How does the amount of freshwater on Earth compare to the amount of saltwater on Earth?

To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i

enyata [817]

Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$ .....(1)

Molarity of acetic acid solution = 0.0880 M

Volume of solution = 30.6 mL

Putting values in equation 1, we get:

$0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol$

Thus there are 0.00269 moles of acetic acid in buffer.

8 0

2 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

3 years ago

A nitrde ion has 7protons,8nutrons and 10 electrons. what is the overall change

Rama09 [41]

Answer:

the overall charge on the nitride anion is

( 3 − ) .

N power 3 − → the nitride anion

6 0

2 years ago

If the distance between two objects is decreased to - of the original

Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

$F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R$