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3 years ago

What is a base used in toothpaste

Chemistry

2 answers:

lakkis [162]3 years ago

8 0

The base of toothpaste is magnesium hydroxide.

AlexFokin [52]3 years ago

6 0

Magnesium hydroxide is the base used in toothpaste. Compounds with hydroxide in their names are known to be basic. A base is used in the toothpaste in order to neutralize the acid produced when we eat

Hope this helped!

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Show a correct numerical setup for calculating the

valkas [14]

0.25 moles of CO2 is present in 11 grams of CO2.

Explanation:

A mole represents the number of chemical entities in an element or molecule.

Number of moles of an element or molecule is determined by the formula:

The Number of moles (n) = weight of the atom given ÷ atomic or molecular weight of the one mole of the element or molecule.

Themolar mass of one mole of carbon dioxide is:

12+ ( 16×2)

= 44 gram/mole

The given weight is 44 grams of carbon dioxide.

Putting the values in the equation,

n= 11 gms÷44 gms/ mole

n = 0.25 mole

7 0

3 years ago

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RoseWind [281]

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3 years ago

Determine the molarity of a solution with a volume of 435. mL and 0.550 mol of solute dissolved.

lora16 [44]

Answer:

M = 1.26

Explanation:

Molarity = mole of solution/liters of solution
435mL/1000 = .435L
Plugging in the numbers into the formula, we get:
Molarity = .550 mol/.435L = 1.26 M

8 0

2 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$