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4 years ago

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A graduated cylinder was filled with water to the 25.0 mL mark and weighed. Its mass was 105.5g. An object made of an unknown me

tal was placed in the cylinder and completely submerged in the water. The water level rose to 30.7 mL. When reweighed, the cylinder, water and metal object had a total mass of 156.0g. Based on these measurement, what is the density of the unknown metal?

1 answer:

Marta_Voda [28]4 years ago

6 0

The answer <span>is <span>8.9 g/mL</span>.</span>

The density (D) is <span>equal to mass (m) divided by volume (V): D = m/V

Let's find the mass of the object:
m = 156 g - 105.5 g = 50.5 g

Let's find the volume of the volume:
V = 30.7 mL - 25 mL = 5.7 mL

The density is:
D = m/V = 50.5 g / 5.7 mL = 8.9 g/mL</span>

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Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

$k_{c}=3.07\times10^{-4}$

Temperature = 24°C

We know that,

The balance equation is

$2NOBr\Rightarrow 2NO+Br_{2}$

Initial concentration is,

$0.878\Rightarrow 0+0$

Concentration is,

$-2x\Rightarrow 2x+x$

Equilibrium concentration

$0.878-2x\Rightarrow 2x+x$

We need to calculate the value of x

Using formula of concentration

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

$2x^2=0.0002367+0.001228x^2-0.0010536x$

$2x^2-0.001228x^2+0.0010536x-0.0002367=0$

$1.998772x^2+0.0010536x-0.0002367=0$

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We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

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