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Gre4nikov [31]
3 years ago
11

Do all metallic oxides form alkaline solution???​

Chemistry
2 answers:
4vir4ik [10]3 years ago
5 0

Answer:

Nope not all

Explanation:

As scientists says that not all give a solution.

sergey [27]3 years ago
5 0
Not all them
hope that helps di
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Zero (0) molecules of glucose are produced.
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How many atoms are in 1.50 moles of Hg?
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Answer:

Atoms=9.033*10^23 atoms

Explanation:

Atoms=no.of miles*Avogadro's no.

Atoms=1.5*6.022*10^23

Atoms=9.033*10^23 atoms

3 0
4 years ago
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Why is elemental zinc considered a pure substance?
Veseljchak [2.6K]

Answer:  C... Because Zinc is made of only one type of atoms.

Explanation: All elements are made of one type of atom. Pure gold is made out of only gold atoms, silver out of only silver atoms, etc.

7 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

4 0
3 years ago
A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in
tamaranim1 [39]

Answer:

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Explanation:

<u>Step 1:</u> Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

<u />

<u>Step 2:</u> Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

<u>Step 3:</u> Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L  * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

<u>Step 4</u>: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

<u>Step 5</u>: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

   

5 0
3 years ago
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