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liq [111]
3 years ago
6

A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de

nsity of the blocking/mL? 1 in = 2.54 cm​
Chemistry
1 answer:
spin [16.1K]3 years ago
4 0

The density of the rectangular block in g/mL is 7.0.

<u>Given the following data:</u>

  • Mass of block = 22.8 gra1.94 kg
  • Length of block = 3.21 cm
  • Width of block = 5.83 cm
  • Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

Volume = length × width × height

<u>Conversion:</u>

1 in = 2.54 cm​

5.83 in = X cm

Cross-multiplying, we have:

X = 2.54(5.83)\\\\X = 14.81 \; cm

Volume = 3.21 × 5.83 × 14.81

Volume = 277.16 cubic centimeters.

<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}

<em>Density </em><em>= 7</em><em>.0 g/mL</em>

Therefore, the density of the rectangular block in g/mL is 7.0.

Read more: brainly.com/question/18320053

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Which of the following will increase the rate of disolving?
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What is the final volume of NaOH solution prepared from 100.0 mL of 0.500 M NaOH if you wanted the final concentration to be 0.1
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Explanation:

We all know that

              V1S1 = V2S2

             or V1= V2S2÷S1

             or  V1= V2×S2×1/S1

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∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.

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To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?
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Answer:

The answer is "6L"

Explanation:

Formula:

\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}

Values:

\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}

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Vanyuwa [196]

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Explanation:

To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.

NOTE: The oxidation number of oxygen (O) is always – 2.

Thus the oxidation number of N in NO can be obtained as follow:

N + O = 0 (ground state)

N + (– 2) = 0

N – 2 = 0

Collect like terms

N = 0 + 2

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Thus, the oxidation number of Nitrogen (N) in NO is +2.

Therefore, the IUPAC name for NO is Nitrogen (ii) oxide

3 0
3 years ago
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