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4 years ago

How many moles are in 74 g KI?

Chemistry

1 answer:

nexus9112 [7]4 years ago

7 0

N=?
m=74g
n=m:Mr
Mr (Cl)=35,5g/mol
n=74g:35,5g/mol
n=2,08mol
I think this is the answer, hope I am right!

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What is the mass of 1.00 mol of oxygen (O2)?

Ad libitum [116K]

Answer:

Mass = 32 g

Explanation:

Given data:

Number of moles of oxygen = 1.0 mol

Mass of oxygen = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of oxygen (O₂) is 32 g/mol

by putting values,

1.0 mol = mass/ 32 g/mol

Mass = 1.0 mol ×32 g/mol

Mass = 32 g

4 0

3 years ago

Please help this will help my grade a lot

vladimir1956 [14]

Answer:

a- customary systems..

3 0

3 years ago

Which of the following types of equations will include spectator ions

krek1111 [17]

Answer:- B) Full ionic equation

Explanations:- A molecular equation does not have ions. It is the non charge form of the elements of compounds. When we write the full or total ionic equation for the molecular equation then ions are written for the aqueous species. The ions present on both sides(reactant and product sides) in a full ionic equation are known as spectator ions or common ions. These ions are canceled to get the net ionic equation. For example, reaction of aqueous solution of silver nitrate with aqueous solution of barium chloride to form aqueous solution of barium nitrate and a precipitate of silver chloride.

Balanced molecular equation:-

$BaCl_2(aq)+2AgNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+2AgCl(s)$

Full ionic equation:-

$Ba^+^2(aq)+2Cl^-(aq)+2Ag^+(aq)+2NO_3^-(aq)\rightarrow Ba^+^2(aq)+2NO_3^-(aq)+2AgCl(s)$

If we look at the above full ionic equation then barium ion and nitrate ions are the spectator ions are they are present on both sides.

So, while writing the net ionic equation, these spectator ions are canceled.

Net ionic equation:-

$2Cl^-(aq)+2Ag^+(aq)\rightarrow2AgCl(s)$

So, it is also clear from the above example that it's the full ionic equation that include spectator ions.

6 0

3 years ago

Read 2 more answers

A laboratory analysis of a 100 g sample finds it is composed of 1.8 g hydrogen, 56.1 g sulfur, and 42.1 g oxygen. What is its em

Neporo4naja [7]

Answer: The empirical formula is $H_2S_2O_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mas of H = 1.8 g

Mass of S = 56.1 g

Mass of O = 42.1 g

Step 1 : convert given masses into moles.

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.8g}{1g/mole}=1.8moles$

Mass of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{56.1g}{32g/mole}=1.8moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.1g}{16g/mole}=2.6moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H = $\frac{1.8}{1.8}=1$

For S = $\frac{1.8}{1.8}=1$

For O = $\frac{2.6}{1.8}=1.5$

Converting to whole number ratios

The ratio of H: S: O= 2: 2: 3

Hence the empirical formula is $H_2S_2O_3$

7 0

3 years ago

snow_tiger [21]

There are 1,000m is 1k. So just move the decimal one position right. 127.56m

There are 10,000cm in 1k. Move the decimal two positions right. 1275.6cm

5 0

4 years ago