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2 years ago

What do scientists Need to know about a compound in order to Classify it

1 answer:

6 0

Compounds are classified according to the elements that make them up. For example, oxides contain one or more oxygen atoms, hydrides contain one or more hydrogen atoms.
compounds form different types of bonds too. a metal and nonmetal element will create an ionic bond, two nonmetal elements create covalent bonds

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Who like math ? Cause I don’t.

seraphim [82]

Answer: it honestly depends.

Explanation:

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2 years ago

Read 2 more answers

Why do A and T (and G and C) pair in a DNA double helix?

Usimov [2.4K]

I just took a quiz on this! It's D!

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2 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

4 0

2 years ago

50 pennies or other coins (all coins must be the same value)

Neko [114]

Hey!

Hope this helps...

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Question 1: Turn 25 coins tails up. You now have 50% parent material and 50% daughter material. What is the half life of this substance? (20 Points)

Answer:

Based upon the given information, we can assume that the half life of this coin material is 5 years.

Question 2: Imagine that another half life has gone by. Remember that ONLY the parent material can change to the daughter material. Turn over your pennies to represent the passing of another half life. How much total parent material and how much total daughter material do you now have? You may express your answer in number of pennies or in %. (Note: Since you cannot turn over half a penny, round the actual number of pennies you would turn over up.) (20 points)

Answer:

Based upon the given information from Question 2 and Question 3, we can assume that 13 additional coins would be turned over. We started with 50, than flipped 50% of them over, and now another 50% over. We have 12 coins left.

Question 3: How many half lives will pass before you have only 3 pennies that are parent material and 47 pennies that are daughter material? (20 Points)

Answer:

Based upon the given information from all the questions, and based upon our answers we can assume this:

1st half-life: 50 turned to 25 coins

2nd half-life: 25 turned to 12 coins

3rd half-life: 12 turns to 6 coins

4th half-life: 6 turns to 3 coins

So, after 4 half-lives (or after 20 years) we would have 3 coins left that have parent material.

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For future reference please do not post as complex questions like this, if you do (post a question that is similar to this) please state the following information:

"All questions posted in this post relate to one-another and cannot be answered without the other questions..."

4 0

2 years ago

Copper crystallizes in a face-centered cubic unit cell. the density of copper is 8.94 g/cm3. calculate the length of the edge of

Ilia_Sergeevich [38]

361.4 pm is the length of the edge of the unit cell. First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight. Atomic weight copper = 63.546 Now divide the mass by the atomic weight, getting 8.94 g / 63.546 g/mol = 0.140685488 mol And multiply by Avogadro's number to get the number of atoms: 0.140685488 * 6.022140857x10^23 = 8.472278233x10^22 Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume. 8.472278233x10^22 / 4 = 2.118069558x10^22 Now calculate the volume each unit cell occupies. 1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3 Let's get the cube root to get the length of an edge. (4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm Now let's convert from cm to pm. 3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.

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2 years ago