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Nady [450]
3 years ago
13

A 4 gram metal object displaces 2 mL of water in a graduated cylinder. What is the density of the metal object?

Chemistry
1 answer:
castortr0y [4]3 years ago
3 0
Density=mass/volume
Density= 4 g/2 mL/cm3
Density =2 g/cm3
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Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite un
maw [93]

Answer:

7.23 J

Explanation:

Step 1: Given data

  • Mass of graphite (m): 566.0 mg
  • Initial temperature: 5.2 °C
  • Final temperature: 23.2 °C
  • Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

6 0
3 years ago
Question: convert 22.63 miles of oxygen to volume<br><br> very urgent
Stels [109]

Answer:

506.912 L

Explanation:

From the question given above, the following data were obtained:

Number of mole of O₂ = 22.63 moles

Volume of O₂ =?

Recall:

1 mole of a gas occupy 22.4 L at STP.

With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.

Thus, 22.63 moles of O₂ is equivalent to 506.912 L.

3 0
3 years ago
In tissue engineering, scaffolds are built for growing replacement body parts primarily so that:
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One of the functions of scaffolds in tissue engineering is letter b. cells grow into the correct shape. Scaffolds used in tissue engineering have biomaterials with biological attributes <span>that will enhance cell attachment </span>and topography where cells morph and align.
7 0
4 years ago
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Which of the following is NOT an example of an insulator?
Korolek [52]
Glass, baecase it dosent insulate, it just collects warmth or coldness.
8 0
3 years ago
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Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0
Alexxx [7]
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2 
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

                                   H2(g) + F2(g) → 2 HF(g)
From reaction        1 mol      1 mol
From problem      2.50 mol   1 .00mol

We can see that  excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

</span>                                 H2(g) + F2(g) → 2 HF(g)
From reaction                      1 mol       2  mol
From problem                      1.00 mol  2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
4 0
3 years ago
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