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Anuta_ua [19.1K]
3 years ago
7

A wooden block has a length of 4.0 cm, a width of 2.0 cm, and a height of 1.0 cm. What is the volume of this block?

Chemistry
1 answer:
aleksklad [387]3 years ago
3 0
Use the formula for volume. V = lwh. Plug in the numbers for length, width, and height. Now we have V = 4.0 x 2.0 x 1.0, which is 8.0. So the volume of the block is 8.0 cm.

Thanks for your question! Don't forget to rate and give me the brainliest answer! Then, I can help you with all your problems! ^-^ ~

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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

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What is the effect of high activation energy on a chemical reaction?
alex41 [277]
<h3><u>Answer;</u></h3>

It makes the reaction harder to start

<h3><u>Explanation</u>;</h3>
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4 0
3 years ago
If a 0.4681 g Mg strip reacts with 0.650 M HCl in a 139.3 mL flask at 25oC, what is the minimum volume (mL) of HCl needed to com
Marizza181 [45]
The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

    Mg + 2HCl --> H2 + MgCl2

The number of moles of HCl that is needed for the reaction is calculated below.
    n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
    n = 0.0385 mols HCl

From the given concentration, we calculate for the required volume. 
    V = 0.0385 mols HCl/(0.650 mols/L)
     V = 0.05926 L or 59.26 mL

<em>Answer: 59.26 mL of HCl</em>
7 0
3 years ago
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