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3 years ago

How do you calculate metal density?

Chemistry

1 answer:

nekit [7.7K]3 years ago

7 0

Answer:

You can divide the mass by the volume to calculate the density of the metal

Explanation:

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Nitromethane (CH3NO2) is an explosive compound sometimes used as a fuel booster in racecars. What is the formal charge of the ni

nevsk [136]

Answer:

Option d. is correct

Explanation:

Nitromethane is an organic compound that can be used as a fuel booster in rockets. Its chemical formula is $CH_3NO_2$ .

This compound is produced as a result of reaction of sodium chloroacetate with sodium nitrite in aqueous solution.

The electron-dot structure of a nitromethane molecule shows $+1$ charge on the nitrogen.

Option d. is correct

5 0

3 years ago

Write the net ionic equation for this reaction occurring in water: ammonium fluoride and magnesium chloride are mixed to form ma

MAXImum [283]

Correct answer: 3. $Mg^{2+}(aq) + 2 F^{-}(aq) --> MgF_{2}(s)$

The given chemical reaction is between ammonium fluoride and magnesium chloride to form magnesium fluoride and ammonium chloride.

The balanced chemical equation representing the reaction will be,

$2NH_{4}F (aq) + MgCl_{2}(aq) -->MgF_{2} (s) + 2NH_{4}Cl (aq)$

The complete ionic equation for the reaction: All the compounds soluble in water (aqueous) will split into ions, $MgF_{2}$ will not split into ions as it is insoluble in water.

$2 NH_{4}^{+}(aq) + 2 F^{-}(aq) + Mg^{2+}(aq) + 2 Cl^{-}(aq) --> MgF_{2}(s) + 2 NH_{4}^{+}(aq) + 2 Cl^{-}(aq)$

The net ionic equation will be:

$Mg^{2+}(aq) + 2 F^{-}(aq) --> MgF_{2}(s)$

Here the spectator ions are $NH_{4}^{+} and Cl^{-}$

6 0

3 years ago

What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N

Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0

4 years ago

14. What are the two main branches of science?

Vlada [557]

Answer:

Physical science

Life science

Explanation:

3 0

3 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (

Taya2010 [7]

Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

NaOH → Na+ + OH-

⇒ C NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ C NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ C HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ C NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ C HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ C HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ C HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

8 0

4 years ago

Read 2 more answers