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Ivahew [28]
3 years ago
12

To make a softball bat, a block of wood is carefully crafted into a shape that will allow a

Chemistry
1 answer:
Iteru [2.4K]3 years ago
4 0
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You might be interested in
Element Q is a theoretical nonmetal with atomic number 59. Consider the isotope: Q-95. How many electrons are in an atom of Q-95
Vinil7 [7]

60

Explanation:

The number of electrons in an atom of Q-95 will be 60 electrons.

The atomic number of an atom is the number of protons in the atom of the isotope

This number is unique to every atom and their isotope

An isotope of an atom is the existence of an atom having the same atomic number but different mass numbers.

Since atomic number the same as number of protons, they are the positively charged particles in an atom.

To have a negative charge, the number of electrons must be more than that of protons.

Charge = number of protons - number electrons

  charge = -1

  -1 = 59  - number of electrons

  Number of electrons = 60 electrons

learn more:

Number of subatomic particles brainly.com/question/2757829

#learnwithBrainly

4 0
3 years ago
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
Need help on dis plz that would be big help
Vaselesa [24]

Answer:

Mg(OH)2

Explanation:

Magnesium hydroxide is the base since it has hydroxyl ions.

4 0
3 years ago
Read 2 more answers
The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?
Mazyrski [523]
Last option that is none of above is right answer.
8 0
4 years ago
Read 2 more answers
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
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