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svlad2 [7]
3 years ago
5

A rock weighing 30.0 g is placed in a graduated cylinder displacing the volume from 22.3 mL to 27.6 mL. What is density of the r

ock in grams/cubic centimeter?
Chemistry
1 answer:
cupoosta [38]3 years ago
5 0
V'=v2-v1=27.6-22.3
density=mass/v'
ps: ml is the same as cubic cm
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A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c
kirill [66]
<span>Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.</span>
3 0
3 years ago
Read 2 more answers
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
3 years ago
How is beryllium similar to calcium?<br>​
Molodets [167]

Answer:

The second column on the periodic table of the chemical elements is collectively called the alkaline earth metal group: beryllium, magnesium, calcium, strontium, barium, and radium. Because the outer electron structure in all of these elements is similar, they all have somewhat similar chemical and physical properties.

Explanation:

8 0
3 years ago
An example of two class b fuels would be
Viefleur [7K]
Flammable liquid,gasoline, oil, and etc
6 0
3 years ago
a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g
leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = 9.77\times \dfrac{22.34}{100}=2.18\ g .

Hence , this is the required solution .

7 0
3 years ago
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