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Elza [17]
3 years ago
13

I really need help with this before 5:10

Chemistry
1 answer:
gtnhenbr [62]3 years ago
4 0
CAN YOU HELP ME ANWSER MINE ILL HELP YOU
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A piece of metal with a mass of 611 g is placed into a graduated cylinder that contains 25.1 ml of water, raising the water leve
jeka94

Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.

Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

V_{metal}=V_{final}-V_{initial}=(56.7-25.1)mL=31.6 mL

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

d=\frac{m}{V}=\frac{611 g}{31.6 mL}=19.33 g/mL

Therefore, density of metal is 19.33 g/mL.

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3 years ago
How does the compound shown in part 5a of the transparency differ from the elements that comprise it?
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How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?
slava [35]

Answer:

A. 0.90 L.

Explanation:

  • NaOH solution will react with H₂SO₄ according to the balanced reaction:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>

<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>

  • For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>

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3 years ago
No explanation needed, just a quick answer please!
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all of the above is the answer :)

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