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4 years ago

When the ph of a solution shifts from 7 to 3, how has the hydrogen ion concentration changed?

1 answer:

7 0

The pH is changing from 7 to 3, which means that the solution will be more acidic because the concentration of H will increase. Remember that pH=-log[H]

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Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Predict the missing product of this equation 1 MgF2 + 1 Li2CO3 -> 1 ______ +2LiF

ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0

3 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0

4 years ago

How is helium different from other noble gases?

Slav-nsk [51]

Helium only possesses two valence electrons, while the other noble gasses posses eight

3 0

3 years ago

Which is the anode and which is the cathode?

rewona [7]

Answer:

2l- ---> l2 + 2e- is the anode

2H+ + 2e- ---> H2(g) is the cathode

Explanation:

Oxidation occurs when a metal loses two or more electrons in a redox chemical reaction and reduction is when it gains. Thus, oxidation is the anode and reduction is the cathode.

5 0

3 years ago