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natali 33 [55]
3 years ago
7

Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign,

calculate the percent error up to two decimal places:Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign, calculate the percent error up to two decimal places:
Chemistry
2 answers:
Marina CMI [18]3 years ago
8 0

Answer:

0.4

Explanation:

The percent error, or relative error, is the error associated with a parameter, and it's the absolute error divided by the parameter.

The absolute error is the differece between the calculated value and the parameter (in module): |2.69 - 2.70| = 0.01 *100% = 1.0%.

The relative error is:

1.0%/2.70 = 0.37% = 0.4%

stepan [7]3 years ago
7 0

Answer:

0.37 %

Explanation:

Given that:

Calculated density of aluminum = 2.69 g/cm³

Accepted density of aluminum = 2.70 g/cm³

Error\ percentage=\frac {|Accepted\ value-Calculated\ value|}{Accepted\ value}\times 100

Thus, applying values as:

Error\ percentage=\frac {|2.70-2.69|}{2.70}\times 100

<u>Percent error = 0.37 %</u>

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Write the symbol for every chemical element that has atomic number greater than 3 and atomic mass less than 15.4 u.
mixas84 [53]

Answer:

N, B, Be, C

Explanation:

N (Nitrogen) has 14.0067 B (Boron) has 10.81 Be (Beryllium) has 9 and C (Carbon) has 12

8 0
2 years ago
What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.
koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

  • 1.000 mol salt is equal to 58.44 g salt
  • 1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

  • 1 L = 1dm³
  • 1 m³ = 10³ dm³

We can use proportions:

10.0m^{3} .\frac{10^{3}dm^{3}  }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt

8 0
3 years ago
An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0
2 years ago
One mole of copper(ii) sulfate, cuso4, contains ________ o atoms.
Ierofanga [76]
According  to  avogadro   constant,  the  number  of  units  in  one  mole  of  any  substance   contain  6.022  x10  ^23  atoms

therefore  the  number  of  o  atoms  in   one  mole  of    CuSO4  =  6.022 x 10 ^ 23
6 0
2 years ago
Read 2 more answers
What is the net ionic equation for the reaction that occurs when aqueous copper(II) sulfate is added to excess 6-molar ammonia?a
Karolina [17]

Answer:

c. 2NH₃ + 2H₂O  + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺

Explanation:

A net ionic equation is a chemical equation that list only the species that are involved in the reaction.

The reaction of ammonia with copper(II) sulfate CuSO₄ in water is:

2NH₃ + 2H₂O  + CuSO₄ → Cu(OH)₂(s) + 2NH₄⁺ + SO₄²⁻

In an ionic equation, salts are written as ions, that means CuSO₄ must be written as Cu²⁺ + SO₄²⁻. That is:

2NH₃ + 2H₂O  + Cu²⁺ +<u> SO₄²⁻</u> → Cu(OH)₂(s) + 2NH₄⁺ + <u>SO₄²⁻</u>

As in a net ionic equation you must list only the species involved in the reaction (The underlined species don't react), the net ionic equation is:

<em>c</em>. <em>2NH₃ + 2H₂O  + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺</em>

<em></em>

I hope it helps!

7 0
3 years ago
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