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avanturin [10]
1 year ago
8

a square boat made from iron has overall dimensions of 2.00 cm x 11.0 cm x 11.0 cm. it has a mass of 213 g. what is its density

242 g/cm3 0.880 g/cm3 51500 g/cm3 1.14 g/cm3
Chemistry
2 answers:
Rus_ich [418]1 year ago
5 0
Volume:

2.00 x 11.0 x 11.0 => 242 cm³

mass : 213 g

D = m / V

D = 213 / 242

D = 0.880 g/cm³

Answer B

hope this helps!
kondor19780726 [428]1 year ago
3 0

Answer: Density of the boat will be 0.880g/cm^3

Explanation: Density of a substance is calculated using the formula:

Density=\frac{Mass}{Volume}   .....(1)

Here, Dimensions of a boat is given, we can calculate the Volume using:

Volume = length × breadth × height

Dimensions are 2.00cm × 11.00cm × 11.00cm

Volume = 2cm × 11cm × 11cm = 242cm^3

Mass of the boat = 213 grams

Putting values in equation 1, we get

Density=\frac{213g}{242cm^3}=0.880g/cm^3

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<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

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Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

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Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

K_a = dissociation constant of an acid = 1.8\times 10^{-5}

K_b = dissociation constant of a base = ?

K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

1.8\times 10^{-5}\times K_b=1\times 10^{-14}

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Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

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pOH=-\log [OH^-]

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