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Anna007 [38]
3 years ago
13

A student reports the density of an article to be 1.92 g/mL. The accepted value of density is 1.89 g/mL. Calculate the student's

percent error.
Chemistry
1 answer:
ipn [44]3 years ago
6 0

Answer:

1.59%.

Explanation:

The following data were obtained from the question:

Measured density = 1.92 g/mL

Accepted density = 1.89 g/mL

Percentage error =...?

The percentage error error of the student can be obtained as follow:

Percentage error = |Measured density – Accepted density |/Accepted density × 100

Percentage error = |1.92 – 1.89|/1.89 × 100

Percentage error = 0.03/1.89 × 100

Percentage error = 1.59%

Therefore, the student percentage error is 1.59%

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Can someone help me check my answers for the mole questions?
Sliva [168]

Answer:

Explanation:

Given data:

Number of particles of phosphorous = 1.2 × 10²⁵

Number of moles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

Number of particles of phosphorous = 1.2 × 10²⁵ particles

1 mole = 6.022 × 10²³ particles

1.2 × 10²⁵ particles × 1 mole /  6.022 × 10²³ particles

0.1993  × 10² mol

19.93 mole

Q 3:

Given data:

Number of moles of calcium = 0.630 mol

Number of particles = ?

Solution:

The number 6.022 × 10²³ is called Avogadro number.

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ particles of hydrogen

For 0.63 moles:

1 mole = 6.022 × 10²³ particles

0.63 mol × 6.022 × 10²³ particles  / 1 mol

3.79 × 10²³ particles

Q 4:

Given data:

Number of moles = 4 mol

Number of particles = ?

Solution:

The number 6.022 × 10²³ is called Avogadro number.

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ particles of hydrogen

For 4 moles:

1 mole = 6.022 × 10²³ particles

4 mol × 6.022 × 10²³ particles  / 1 mol

24 × 10²³ particles

2.4  × 10²⁴ particles

3 0
3 years ago
Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. Th
Sedaia [141]

Answer:

154 g

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

79.5 L × 1 mol/22.4 L = 3.55 mol

Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.

Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.37 mol × 65.01 g/mol = 154 g

6 0
3 years ago
PLEASE HELP!!!!! NEED HELP ASAP!!!!!
Anarel [89]

Answer:

1.BaCO3 2.MgBr2 3.aluminum and oxygen 4.Potassium chloride

Explanation:

3 0
2 years ago
If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?
MrMuchimi

We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.

The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.

Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, \frac{1 X 0.75}{2} moles = 0.375 moles of nitrogen is required.

Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.

Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.


5 0
3 years ago
How much water vapor can a cubic meter of air hold at 25 degrees celcius
Thepotemich [5.8K]
I think you might find your answer here https://answers.yahoo.com/question/index?qid=20121101101721AAeZyHE
5 0
3 years ago
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