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Mekhanik
3 days ago
8

For the chemical reaction HCN ( aq ) + KOH ( aq ) ⟶ H 2 O ( l ) + KCN ( aq ) HCN(aq)+KOH(aq)⟶H2O(l)+KCN(aq) write the net ionic

equation, including the phases. net ionic equation:
Chemistry
1 answer:
lions [985]3 days ago
4 0

Answer:

H+(aq) + OH-(aq) ⟶ H2O(l)

Explanation:

Step 1: The balanced equation can be presented as

HCN(aq) + KOH(aq) ⟶ H2O(l) + KCN(aq)

H+(aq) + CN-(aq) + K+(aq) + OH-(aq) ⟶ H2O(l) + K+(aq) + CN-(aq)

Step 2: The net ionic equation is formed by removing the spectator ions — those that appear on both sides of the equation — leading us to the final representation:

H+(aq) + OH-(aq) ⟶ H2O(l)

You might be interested in
7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.
lorasvet [946]

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

7 0
10 days ago
Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [845]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

6 0
17 days ago
Read 2 more answers
You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di
lorasvet [946]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

500mM = 500mM \times (\frac{1M}{1000M})= 0.5M

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M

10 mL of this solution is further diluted to 250 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M

Convert μM:

2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M

5 mL of this solution is diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M

Convert μM:

0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M

10 mL of this solution is further diluted to 500 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

7 0
9 days ago
A certain liquid has a density of 2.67 g/ cm3. what is the mass of 30.5 ml of this liquid? (
lorasvet [946]
Hello!

density = 2.67 g/cm³

volume = 30.5 mL

Thus:

Mass = density * volume

Mass = 2.67 * 30.5

Mass = 81.435 g 
4 0
6 days ago
What is the volume in liters of a cube whose edge is 4.33 cm long?
VMariaS [1037]
Assuming the edge length of the cube is m.
Since the cube's sides are all equal, you can find its volume using the formula:
volume of cube = m^3

Now, using the provided edge length:
volume = (4.33)^3 = 81.1827 cm^3

According to unit conversion standards, 1 cm^3 = 0.001 liters.
So:
volume of cube = 81.1827 cm^3 = 0.0811827 liters
5 0
19 hours ago
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