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How many moles of calcium, Ca, are in 5.00 g of calcium?

Rom4ik [11]

<h3>Answer:</h3>

0.125 mol Ca

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

5.00 g Ca

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

Molar mass of Ca - 40.08 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 5.00 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca})$
Multiply: $\displaystyle 0.12475 \ mol \ Ca$

<u>Step 4: Check</u>

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

0.12475 mol Ca ≈ 0.125 mol Ca

3 0

2 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

2 years ago

If the solubility of sodium chloride is 36 grams per 100 grams of water, which of the following solutions would be considered un

Murrr4er [49]

Answer: If the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

Explanation:

A solution which contains the maximum amount of solute is called a saturated solution. Whereas a solution in which more amount of solute is able to dissolve is called an unsaturated solution.

Now, the number of moles present in 36 g of NaCl (molar mass = 58.4 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{36 g}{58.4 g/mol}\\= 0.616 mol$

This shows that solubility of sodium chloride is 36 grams per 100 grams of water means a maximum of 0.616 mol of NaCl will dissolve in 100 mL of water.

So, a solution in which number of moles of NaCl are less than 0.616 mol per 100 mL then the solution formed will be an unsaturated solution.

As 5.8 moles of NaCl dissolved in 1 L (or 1000 mL) of water. So, moles present in 100 mL are calculated as follows.

$Moles = \frac{5.8 mol}{1000 mL} \times 100 mL\\= 0.58 mol$

Moles present in 100 mL of water for 3.25 moles of NaCl dissolved in 500 ml in water are as follows.

$Moles = \frac{3.25 mol}{500 mL} \times 100 mL\\0.65 mol$

Moles present in 100 mL of water for 1.85 moles of NaCl dissolved in 300 ml of water are as follows.

$Moles = \frac{1.85 mol}{300 mL} \times 100 mL\\= 0.616 mol$

Thus, we can conclude that if the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

8 0

3 years ago

Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb

Vikki [24]

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer

8 0

3 years ago

Read 2 more answers

Why is it incorrect to call an astronaut massless?

Reil [10]

<span>Why is it incorrect to call an astronaut massless?

its incorrect because everything has mass wheather in space or on a planet.</span>

7 0

3 years ago

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