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jeka94
3 years ago
12

Chloroform has a density of 1.483g/ml. A sample of chloroform has a volume of 1.93 liters. How many kilograms does it weight? ​

Chemistry
1 answer:
ollegr [7]3 years ago
8 0

Weight of Chloroform : = 2.862 kg

<h3>Further explanation</h3>

Given

Density 1.483 g/ml

Volume = 1.93 L

Required

Weight of Chloroform

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Convert density to kg/L :

=1.483g/ml = 1.483 kg/L

So the weight(mass) :

= ρ x V

= 1.483 kg/L x 1.93 L

= 2.862 kg

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Near Tatooine's famous Mos Eisley Space port you found a 520.79 mL block of gold. What is the mass in kilograms of this block?
postnew [5]

Answer:

10.051kg of gold is the mass in kilograms

Explanation:

The density of gold is 19.3g/mL. To solve this question we need to convert the volume of the block of gold to grams using its density. Then, we must convert grams to kilograms using the equivalence factor (1000g = 1kg):

<em>Mass in grams:</em>

520.79mL * (19.3g / mL) = 10051g of gold

<em>Mass in kilograms:</em>

10051g of gold * (1kg / 1000g) = 10.051kg of gold

3 0
3 years ago
Iron (Fe) has three isotopes. Calculate the average atomic mass of the element Fe using the following data: Isotope atomic mass
Maksim231197 [3]

Answer:

Average atomic mass = 55.83 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})

Given that:

For first isotope, Fe-54:

% = 6 %

Mass = 53.94 amu

For second isotope, Fe-56:

% = 92 %

Mass = 55.93 amu

For third isotope, Fe-57:

% = 2 %

Mass = 56.94 amu

Thus,  

Average\ atomic\ mass=\frac{6}{100}\times {53.94}+\frac{92}{100}\times {55.93}+\frac{2}{100}\times {56.94}

<u>Average atomic mass = 55.83 amu</u>

5 0
3 years ago
Aluminum metal reacts with dilute sulfuric acid to produce aluminum sulfate and hydrogen gas. what mass of aluminum will react w
SOVA2 [1]
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2

From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams

Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams

From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid

To get the amount  of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
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3 0
3 years ago
The highest point on a wave is the ______ while the lowest point is the ____
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The highest point = crest

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4 0
3 years ago
A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6
Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

Finally, the specific heat of this metal is

H = 993 \frac{J}{kgC}

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally,  we can conclude that this metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

7 0
3 years ago
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