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Alex_Xolod [135]
3 years ago
11

how do you overcome or reduce the problem of random error and systematic error while doing experiment

Chemistry
1 answer:
Ulleksa [173]3 years ago
8 0
<span> The best way to overcome or reduce the problem of random error and systematic error while doing experiment is to "increase the sample size" </span>
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2. Food is another concern before the trip begins. Everyone has become accustomed to a 1500 calorie diet due the food
fomenos

About 46.1 pounds of apple would be needed per day of the trip.

I managed to extract the complete scenario from the attached image.

The trip is made up of 4 men and 3 women, making a total of 7 people.

Everyone has become accustomed to 1500 calories per day.

The total calories needed per day for 7 persons = 1500 x 7

                           = 10,500 calories

189 g of apple has only 95 calories, what g of apple will produce 10,500 calories?

                 = 10,500 x 189/95 = 20,889.67 g of apple

From conversion:

1 g = 0.00220462 pounds

20,889.67 g = 20,889.67 x 0.00220462

                         = 46.1 pounds of apple

More about dimensional analysis can be found here: brainly.com/question/22015862

8 0
3 years ago
When atoms from Column I (Group 1) combine with atoms from Column VII (Group 17): A nearly 100% covalent bond forms. The bond wi
gtnhenbr [62]

The answer you're looking for is B. The bond will be ionic.

7 0
3 years ago
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A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If t
11Alexandr11 [23.1K]

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

8 0
3 years ago
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Would be glad if you helped me!
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b) <u>Cu⁺² + 2e⁻ -----> Cu⁰  reduction      |*3</u>

c) 2Fe⁰ +3Cu⁺² -----> 2Fe⁺³ + 2Cu⁰
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3 years ago
How many grams of iodine are needed to prepare 28.6 grams of ICl
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The answer will be 17.9 grams
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