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Usimov [2.4K]
3 years ago
12

A 68-gram bar of gold is cut into three equal pieces. How dose the density of each piece compare to the density of the original

gold bar?
Chemistry
1 answer:
Molodets [167]3 years ago
8 0
It is the same density
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Does the model below represent a mixture or a substance ?
Vikentia [17]

Answer:

substance

Explanation:

A mixture is when two or more <u>different</u> atoms/molecules are together, but not joined.

A substance is when the <u>same </u>atom/molecule is in a group together.

In this example, it is a substance because it is comprised of the same molecule not joined all together. If you wanted a mixture, other colored atoms/molecule (e.g. add green atoms) would change it to this property.

4 0
3 years ago
How many grams of NaCl are in 2.0 Liters of a 3.0 M NaCl solution?
Naddik [55]

Answer:

350.64g

Explanation:

So first you must know that M is mol/L

Next solve the problem using dimensional analysis

2L NaCl (3 mol/L) = 6 mol NaCl

After you got the number of moles you should look at your periodic table to find the molar mass

I see that it's 58.44g/mol

Use dimensional analysis again!

6 mol (58.44g/mol) = 350.64g

Don't forget to make me brainliest!

4 0
3 years ago
In Mitosis, do new cells have different DNA or the same DNA?
Otrada [13]

Answer: With few exceptions, the mitotic process ensures that this is the case. Therefore, mitosis ensures that each successive cellular generation has the same genetic composition as the previous generation, as well as an identical chromosome set.

5 0
3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
3 years ago
What is the total mass of oxygen contained in 8.20g of benzoic acid (C6H5COOH)
VashaNatasha [74]

Answer:

Mass of oxygen = 2.2 g

Explanation:

Given data:

Mass of benzoic acid= 8.20 g

Mass of oxygen= ?

Solution:

Molar mass of oxygen = 16×2 g/mol

Molar mass of C₆H₅COOH = 7×12 + 1×6 + 2×16

Molar mass of C₆H₅COOH = 84 + 6 + 32

Molar mass of C₆H₅COOH = 122g/mol

Mass of oxygen in 8.20 g of C₆H₅COOH :

Mass of oxygen = 32 g.mol⁻¹/122 g.mol⁻¹ × 8.20 g

Mass of oxygen = 2.2g

4 0
3 years ago
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