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3 years ago

A metal, M, forms an oxide compound with the

Chemistry

1 answer:

lapo4ka [179]3 years ago

5 0

Answer:

Group 1

Explanation: because the compound has a formula of M2O , the number of valence electrons of M should be 1.

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Asap... define pure substances...

gtnhenbr [62]

A pure substance refers to an element or a compound that has no component of another compound or element. Pure substances are made of only one type of atom or molecule. Hydrogen gas and pure iron are examples of pure substances. Hydrogen consists of hydrogen atoms only while iron consists of only iron atoms. Mixing two pure substances results in a mixture. To separate the two, scientists use a method known as filtration. Mixtures can either be homogeneous or heterogeneous. The measure used to determine how pure a substance may be called purity. Besides hydrogen and iron, other pure substances include gold, diamonds, sugar, and baking soda.

3 0

3 years ago

Read 2 more answers

How many atoms are there in 3.559*10^-6 mol of krypton?

algol [13]

Answer:

The answer is

<h2> $2.143 \times {10}^{18} \: \: atoms$ </h2>

Explanation:

To find the number of atoms given the number of moles we use the formula

N = n × L

where

N is the number of entities

n is the number of moles

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

$n = 3.559 \times {10}^{ - 6} \: mol$

Substitute the values into the above formula and solve

That's

<h3> $N = 3.559 \times {10}^{ - 6} \times 6.02 \times {10}^{23} \\ = 2.1425 \times {10}^{18}$ </h3>

We have the final answer as

<h3> $N = 2.143 \times {10}^{18} \: \: atoms$ </h3>

Hope this helps you

3 0

3 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$