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3 years ago

What is the density of a sample of rubbing alcohol if it has a specific gravity of 0.789?

1 answer:

ioda3 years ago

6 0

Specific gravity is a dimensionless unit defined as the ratio of the density of a substance to the density of water - at a specified temperature and can be expressed as. SG = ρsubstance / ρH2O . We calculate as follows:

0.789 = ρsubstance / 1000 kg/m^3
ρsubstance = 789 kg / m^3

Hope this answers the question. Have a nice day.

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In a solution the solute is the substance present in the greatest amount true or false

kati45 [8]

Solute is something that is being dissolved { ex : sugar , salt}

Solvent is something that has ability to dissolve things { ex : water}

False because the solvent is present in larger amounts...

3 0

4 years ago

Read 2 more answers

1. A student collected the following data for a fixed volume of gas: Temperature (⁰C) Pressure (mm of Hg) 10 726 20 750 40 800 7

zimovet [89]

Answer is: the missing pressure is 1088.66 mmHg.

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 960 mmHg; pressure of the gas.

T₁ = 100°C + 273.15.

T₁ = 373.15 K; temperature of the gas.

T₂ = 150°C + 273.15.

T₂ = 423.15 K.

p₂ = p₁T₂/T₁.

p₂ = 960 mmHg · 423.15 K / 373.15 K.

p₂ = 1088.66 mmHg.

6 0

4 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

6 0

3 years ago

At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

Answer: The equilibrium concentration of bromine gas is 0.00135 M

Explanation:

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

Initial: 0.02 0.03

At eqllm: 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}$

We are given:

$K_c=0.141$

Putting values in above equation, we get:

$0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135$

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

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3 years ago

Students in a science class were studying the physical properties of various elements and compounds. One lab group observed that

katrin2010 [14]

Answer:

A:metal

Explanation:

because if a light bulb was placed on metal with force it would light

7 0

3 years ago