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zlopas [31]
3 years ago
12

Three students use a meterstick with millimeter markings to measure a length of wire. Their measurements are 3 cm, 3.3 cm, and 2

.87 cm,
respectively. Explain which answer was recorded correctly.
Chemistry
2 answers:
Volgvan3 years ago
7 0

Answer: 3 cm

Explanation:

andreev551 [17]3 years ago
3 0
The answer is 3 I had this two
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Converting 53.3g of oxygen to moles will give you???
Marat540 [252]

Answer:

3.33 mol

Explanation:

1 g of oxygen is 0.062502343837894 mol

53.3 * 0.062502343837894

3.3313749265597505 mol

8 0
3 years ago
If an electrostatically charged object is placed near other objects, which of the following will occur? Select all that apply. I
Alja [10]

Answer:

As the electrostatically charged object is to be placed in the field of charged particles it will be attracted to those who would be of oppositely charged and repelled by the same charged particles. phenomenon of like charges repel and opposite charges attract each other will be carried out and no deflection will be shown by the charge towards the neutral charge.

5 0
3 years ago
Read 2 more answers
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
In comparing a 0.25 molality aqueous NaCl solution to a 0.25 molality aqueous CaCl 2 solution:
4vir4ik [10]

Answer:

d

Explanation:

higher no of particles are in CaCl2 solution

4 0
2 years ago
Which of these is NOT true about electromagnets?
Alex17521 [72]

Answer:

Permanent = false

Explanation:

All of the other choices are true

4 0
2 years ago
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