James adds some magnetic marbles to a glass jar full of ordinary marbles, and then shakes up the jar.

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Answer 1

Answer:

Answer: Magnetic marbles will tend to attract each other

Related Questions

QUICK WILL MARK BRAINLIEST!!Match each of the unknown ions to its appropriate description.A− A) A nonmetal that gained one electronB+ B) A metal that lost one electronC2− C) A metal that lost two electrons D2+ D) A nonmetal that gained two electrons
Round each answer to the number of significant figures in the parentheses.
Perform the calculation and report the answer using the proper number of significant figures. Make sure the answer is rounded correctly. 1.012×10^-3 J/(0.015456 g)(298.3682−298.3567)K
Combustion of Solid Fuel. A fuel analyzes 74.0 wt % C and 12.0% ash (inert). Air is added to burn the fuel, producing a flue gas of 12.4% CO2, 1.2% CO, 5.7% O2, and 80.7% N2. Calculate the kg of fuel used for 100 kg mol of outlet flue gas and the kg mol of air used. (Hint: First calculate the mol O2 added in the air, using the fact that the N2 in the flue gas equals the N2 added in the air. Then make a carbon balance to obtain the total moles of C added.)
Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

Weathering and erosion can turn metamorphic rock into sediment. Which of the following processes would turn that sediment into sedimentary rock? * 10 points weathering and erosion compaction and cementation heat and pressure melting

Answers

Answer:

compaction and cementation

Explanation:

sedimentary rock is layered rock, in order to become sedimentary it most be compacted to form layers and cemented to become hard. This is what i think. Hope i can help :)

Which three are advantages of asexual reproduction?A:Offspring are more likely to survive environmental changes.

B:Some offspring are more likely to survive a disease.

C:Less energy is required to reproduce.

D:The population can increase from only one parent.

E:The population can increase quickly.

Answers

C, D, and E

A and B cannot be true because asexual reproduction means the parent organism is essentially creating clones of itself, providing no variation in DNA and making all offspring vulnerable to the same environmental changes and diseases as the parent.

Answer:

sorry if I get this wrong I think it is C

Explanation:

One cubic meter of an ideal gas at 600 K and 1000 KPa expand to five times its initial volume as follows:(a) By a mechanically reversible, isothermal process [T 2 = 600K; P 2 = 200kPa; W= -1609 kJ]

(b) By a mechanically reversible, adiabatic process [T 2 =208.96K; P 2 = 67.65 kPa; W= -994.4 kJ] For each case calculate the final temperature, pressure and the work done by the gas. C p =21 Jmol- 1K-1.

Answers

Answer:

lol just grabbing your points

Explanation:lol just grabbing your points

lol just grabbing your points

How many formula units are in a 7.3 x 10-3 gram sample of lithium chloride, LiCI?

Answers

Answer: $1.0 * 10^(20)$ formula units

Explanation:

so lets start off by looking at what we have. we have $7.3 * 10^(-3)$ g of LiCl

which is called Lithium Chloride. in order to convert g to moles, we divide the g by the molar mass of Lithium Chloride.

whip out that HANDY DANDY PERIODIC TABLE man the PERIODIC TABLE WILL SAVE YOUR LIFE SOME DAY! someone will walk up to you all mean, and youll be like, "what, you tryna MUG me?" and then you whack 'em with the periodic table like BAM! GOTTEM!

okay so lets look at the periodic table and we notice that the atomic mass of Lithium is 6.941 and the atomic mass of Chlorine is 35.453. notice that in LiCl there is only one of each. so lets add 6.941 + 35.453 = 42.394 g/mol.

now look at what we were given: converting the given quantity to standard format instead of scientific format, we have 0.0073 grams of lithium chloride. we can convert this to moles by dividing it by its molar mass which is 42.394. $0.0073 / 42.394 = 1.72 * 10^(-4)$

now lets use AVOCADOS NUMBER i mean AVOGADROS NUMBER!! which is $6.02 * 10^(23)$

multiply $(1.72 * 10^(-4)) * (6.02 * 10^(23))$ and we get $10.35 * 10^(19)$ formula units.

if you want to be specific about the significant figures, notice that the given quantity in the question only has two significant figures. so we can alter our final answer to only have two sig figs. lets change it: $1.0 * 10^(20)$ formula units

A laboratory scientist wants to analyze the chlorogenic acid content of a 0.5 L sample. He uses an indicator solution and titrates with sodium hydroxide. He finds the equivalence point occurs when he's titrated 9.5 mL of 0.05 M sodium hydroxide. Approximately how much chlorogenic acid is in the sample? 475 umol ® 480 umol © 500 umol 510 umol

Answers

Answer:

There are approximately 475 umol of chlorogenic acid in the sample.

Explanation:

The first step is indentifying the chlorogenic acid structure. As it can be seen in the figure attached, this molecule is a carboxylic acid containing just one carboxyl group. This means, that chlorogenic acid is a monoprotic acid and it is only able to donate one proton per molecule or one mol of protons per one mol of molecules.

The second step is to balance the titration equation. Considering that sodium hydroxide will generate one mol of hydroxyl ions per mol of salt, we can simplify the equation:

H⁺ + OH ⁻ → H₂O

Therefore, we now know that for each mol of NaOH consumed 1 mol of chlorogenic acid is titrated.

Thus, the last step is calculation the amount of NaOH consumed during the tritation. We can use the following equation:

$C = (n)/(V)$

In which C is the concentration, n the amount of moles and V the volume.

$0,05 M = (n)/(0.0095 L)$

The result is that n = 475 umol.

Need help I don't remember what to do?

Answers

I'm just doing the ones that you don't have numbers already for.
2.) just leave it alone and it's correct
3.) Mg + 2AgNo3 --> Mg(No3)2 + 2Ag
5.) just leave it alone and it's correct
8.) 10C3H8O3 + 15O2 --> 30CO2 + 4H2O
10.) P4 + 6Br2 --> 4PBr3
12.) 2FeCl3 + 6NaOH --> 2Fe(OH)3 + 6NaCl
13.) 2CH3OH + 3O2 --> 2CO2 + 4H2O
14.) 2Al + 3Cu(NO3)2 --> 2Al(NO3)3 + 3Cu
15.) 3CaCl2 + 2K3AsO4 --> Ca3(AsO4)2 + 6KCl
16.) 2NH3 --> N2 + 3H2
17.) 2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
19.) Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
I hope this helps you!!

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