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From a 2.875 g sample containing only iron, sand, and salt, 0.660 g of iron and 1.161 g of sand were separated and recovered. What was the percent of salt in the original sample

Respuesta :

dsdrajlin

Answer:

36.66%

Explanation:

Step 1: Given data

  • Mass of iron: 0.660 g
  • Mass of sand: 1.161 g
  • Mass of the sample: 2.875 g
  • Mass of salt: ?

Step 2: Calculate the mass of salt

The mass of the sample is equal to the sum of the masses of the components.

m(sample) = m(iron) + m(sand) + m(salt)

m(salt) = m(sample) - m(iron) - m(sand)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 3: Calculate the percent of salt in the sample

We will use the following expression.

%(salt) = m(salt) / m(sample) × 100%

%(salt) = 1.054 g / 2.875 g × 100% = 36.66%

mahima6824soni

The percent of salt in the original sample containing only iron, sand, and salt is 36.66%.

What is salt?

Salt is a mineral made up of primarily sodium chloride.

Seawater contains a vast amount of salt.

Given,

The total mass of the sample is 2.875 g

Mass of iron is  0.660 g

Mass of sand is 1.161 g

Step 1: To find the mass of salt

The total mass of the sample = the sum of the masses of the compounds

m(sample) = m(iron) + m(sand) + m(salt)

2.875 g =  0.660 g + 1.161 g + m(salt)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 2: Calculate the percent of salt in the sample

[tex]\%(salt) = \dfrac{m(salt) }{m(sample)} \times 100\%\\\\\\\%(salt) = \dfrac{1.054 g }{2.875 g} \times 100\% = 36.66\%[/tex]

Thus, the percentage of salt in the sample is 36.66%.

Learn more about salts, here:

https://brainly.com/question/16446935

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