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3 years ago

When chlorine and magnesium react chemically, how many chlorine atoms will combine with one magnesium atom?

1 answer:

8 0

Magnesium is divalent. This means that one magnesium atom needs to lose 2 electrons in order to become stable.

Chlorine, on the other hand, is monovalent. This means that one chlorine atom needs to gain one electron in order to become stable.

Based on this, one magnesium atom will combine with two chlorine atoms, where the magnesium loses two electrons, one for each chlorine.

The formula of the compound formed is: MgCl2

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An infant ibuprofen suspension contains 100 mg/5.0mL suspension. The recommended dose is 10 mg/kg body weight.How many millilite

Molodets [167]

Converts the 25 lb to kg

multiply the kilograms by the dose (10 mg/kg)

multiply the amount of ibuprofen by the conversion factor of the concentration of the suspension to calculate the mL

25 lb x 0.453592 kg/1lb= "A kg"

A kg x 10 mg/kg = B mg

B mg x 5.0 mL/ 100 mg = "C mL of suspension"

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3 years ago

Identify a chemical reagent used in this experiment that can be used to distinguish solid CaCl2 (soluble) from solid CaCO3 (inso

stiks02 [169]

A chemical reagent that is used in this experiment is silver nitrate (AgNO3). It is used to distinguish calcium chloride and calcium carbonate. when this reagent is used, silver from silver nitrate reacts with Chloride to calcium chloride and forms silver chloride, making a precipitates of white color.

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3 years ago

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Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 8.00 g of octane is m

DerKrebs [107]

Answer:

11.3 g of H₂O will be produced.

Explanation:

The combustion is:

2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O

First of all, we determine the moles of the reactants in order to find out the limiting reactant.

8 g / 114g/mol = 0.0701 moles of octane

37g / 32 g/mol = 1.15 moles of oxygen

The limiting reagent is the octane. Let's see it by this rule of three:

25 moles of oxygen react to 2 moles of octane so

1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.

We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:

2 moles of octane produce 18 moles of water

Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.

We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.

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3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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4 years ago

Am I weird or is no one seeing my qn<br>free point by the way

Julli [10]

Answer:

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Explanation:

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2 years ago

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