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1 year ago

what salt is formed in the reaction of aluminum with hydrochloric acid? write the formula of the salt. formula:

1 answer:

3 0

Aluminum chloride is formed in the reaction of aluminum with hydrochloric acid. The formula of aluminum chloride is AlCl₃.

Aluminium chloride, additionally referred to as Aluminium trichloride, is an inorganic compound with the formula AlCl₃. It bureaucracy hexahydrate with the system [Al(H₂O)₆]Cl₃, containing six water molecules of hydration. Each are colourless crystals, however samples are regularly contaminated with iron(III) chloride, giving a yellow shade.

Aluminum Chloride is used to control immoderate sweating. This medicinal drug can be used for other purposes; ask your fitness care provider or pharmacist when you have questions.

Aluminum chloride is corrosive and nerve-racking to the eyes, skin, and mucous membranes. Can be dangerous if swallowed. Ingestion of big quantities may also motive phosphate deficiency. Aluminum chloride is a compound of aluminum and chloride it truly is widely utilized in petroleum refining and the producing of many products. similarly, antiperspirants and cosmetic astringents use this compound.

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Net ionic equation of hydrated oxalic acid and sodium hydroxide

blondinia [14]

6 0

3 years ago

675 g of carbon tetrabromide is equivalent to how many

VARVARA [1.3K]

<h3>Answer:</h3>

2.04 mol CBr₄

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Organic</u>

Writing Organic Compounds
Writing Covalent Compounds
Organic Prefixes

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

675 g CBr₄

<u>Step 2: Identify Conversions</u>

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

<u>Step 3: Convert</u>

<u /> $\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4$ <u />

<u />

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

7 0

2 years ago

Read 2 more answers

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

2 years ago

You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b

alexdok [17]

Answer:

0.045 L or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

8 0

2 years ago

M/ABCI<br>m.CADC = 112<br>BK D 117

raketka [301]

Answer:

what do u mean????? i dont get the question

4 0

3 years ago