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BaLLatris [955]
2 years ago
3

Scale test mineral 10 diamond 9 corundum 8 topaz 7 quartz 6 feldspar 5 apatite 4 fluorite 3 calcite 2 gypsum 1 talc mineral h. d

ensity, g/cc tellurium 2 2.07 galena 2.5 7.58 anglesite 2.5 - 3 6.4 chalcocite 2.5 - 3 5.6 copper 2.5 - 3 9.0 gold 2.5 - 3 19.3 silver 2.5 - 3 10.5 arsenic 3.5 5.7 barite 3 - 3.5 4.4 dolomite 3.5 - 4 2.9 platinum 4.5 21.5 willemite 5.5 4.0 magnetite 6 5.18 pyrite 6 - 6.5 5.02 pyrolusite 6 - 6.5 5.0 cassiterite 6.5 6.9 diamond 10 3.52 volume of water displaced = 0.175 l mass: g density. g/cm³ hardness: = - determine the unknown mineral. unknown mineral:
Chemistry
2 answers:
Drupady [299]2 years ago
7 0

Answer:

Mass: 981.0 g

Density: 5.61 g/cm^3

Hardness: 2.5-3

Unknown Mineral: chalcocite

hoa [83]2 years ago
3 0

im pretty sure the answer is chalcocite based off of the hardness, and the options given that the one that fits best

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Write a balanced half-reaction describing the reduction of aqueous vanadium(V) cations to aqueous vanadium(I) cations.
olga nikolaevna [1]

Reduction refers to a chemical reaction in which electrons are gained by one of the atoms involved in the reaction and at the same time lowering of an oxidation state of that atom takes place.  

In the given case, the reduction occurs at the cathode. In aqueous, V (vanadium) is present in +5 oxidation state that on reduction modifies to vanadium (aq) with +1 oxidation state.  

The half reaction is:  

V⁵⁺ (aq) + 4e⁻ → V¹⁺ (aq)


7 0
2 years ago
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A particular car has a gas mileage of 24.5 miles per gallon. If the cost of gas is $4.25 per gallon, and the car travels at a co
tester [92]

Answer:

$30.39

Explanation:

alot of math

3 0
2 years ago
Type the correct answer in the box. Express your answer to three significant figures.
VladimirAG [237]

<u>Given:</u>

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

<u>To determine:</u>

Theoretical yield of calcium phosphate, Ca3(PO4)2

<u>Explanation:</u>

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 =  96.1 g

Molar mass of Ca(NO3)2 =  164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g



7 0
2 years ago
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2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.
jolli1 [7]

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, 0.5^{n}

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

8 0
2 years ago
Read 2 more answers
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