All living things are made of organic compounds which contain the element

Which reagent is the limiting reactant when 0.700 mol al(oh)3 and 0.700 mol h2so4 are allowed to react?

lisov135 [29]

Hffgbhv vice uvula Gn kick kill is Lal Lal lam Lal lisp loco num Mann ight isms genus in I icon 25

3 0

3 years ago

Given the following unbalanced equation:

antiseptic1488 [7]

<h3>Answer:</h3>

11.84 mol CoF₂

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

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

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 11.84 \ mol \ CoF_2$

7 0

2 years ago

How many moles of chlorine (Cl) atoms are in a sample of 1.72 × 1022 atoms? 0.0286 mol Cl 35.0 mol Cl 1.03 × 1023 mol Cl 1.04 ×

kogti [31]

Given information : Sample of $1.72\times 10^{22} atoms$

We need to find the moles of Chlorine in the given sample.

We can say that we need to find moles from the given atoms.

Relation between mole and atom is given by : $1 mole = 6.022\times 10^{23} atoms$

Where $6.022\times 10^{23}$ is Avogadro number.

$1.72\times 10^{22} atoms\times \frac{1 mole}{6.022\times 10^{23} atoms}$

$1.72\times 10^{22}\times \frac{1 mole}{6.022\times 10^{23}}$

On solving the above equation , atoms(unit) gets cancelled out and we get 0.0286 mol.

In the given sample the moles of Chlorine (Cl) is 0.0286 mol , so option A is correct.

6 0

3 years ago

Find the value of the solubility product ksp for a saturated solution of mercury bromide, hgbr2, using the solubility of mercury

kondaur [170]

When the reaction equation is:

HgBr2(s) ↔ Hg2+(aq) + 2Br-(aq)

So Ksp expression = [Hg2+] [Br-]^2

assume the solubility S = X = 2.66 x 10^-7 M

and from the reaction equation :

we can see that [Hg2+] = X

and the [Br-] = 2 X

so by substitution in Ksp formula will can get the Ksp value:

∴ Ksp = X * (2X)^2

= 2.66 x 10^-7 * (2*2.66 x 10^-7)^2

= 7.53 x 10^-20

7 0

3 years ago

The Synthesis Reaction of aluminum (Al) and iodine (I) forms a new, more complex

marysya [2.9K]

Answer:

Aluminum iodide (AlI₃)

Explanation:

The synthesis reaction of aluminum (Al) and iodine (I) can be illustrated as shown below:

Aluminium exhibit trivalent positive ion (Al³⁺)

Iodine exhibit univalent negative ion (I¯)

During reaction, there will be an exchange of ion as shown below:

Al³⁺ + I¯ —> AlI₃

Thus, we can write the balanced equation for the reaction as follow:

Al + I₂ —› AlI₃

There are 2 atoms of I on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of AlI₃ and 3 in front of I₂ as shown below:

Al + 3I₂ —› 2AlI₃

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al + 3I₂ —› 2AlI₃

Thus the equation is balanced.

The product on the reaction is aluminum iodide (AlI₃)

3 0

3 years ago