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Sindrei [870]
2 years ago
9

A CHM 111 student completing Part B recorded the mass of several pieces of dry metal as

Chemistry
1 answer:
Leokris [45]2 years ago
4 0

The density of the metal in units of grams per milliliter is 11.31 g/mL

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to determine the density of the metal </h3>

The following data were obtained from the question:

  • Mass of metal = 23.7412 g
  • Volume of water = 25 mL
  • Volume of water + metal = 27.1 mL
  • Volume of metal = 27.1 – 25 = 2.1 mL
  • Density of metal =?

Density = mass / volume

Density of metal = 23.7412 / 2.1

Density of metal = 11.31 g/mL

Learn more about density:

brainly.com/question/952755

#SPJ1

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wariber [46]

Answer:

The second one is the answer

4 0
3 years ago
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Did I get This right I’m confused
Paul [167]
Hey bud !
You were close congrats !
the only problem i saw was the units part at the end.
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3 0
4 years ago
If a chemical reaction goes to completion, what can we say about the equilibrium constant for that reaction?
aniked [119]

Answer:

A. The equilibrium constant is very large

Explanation:

The equilibrium constant value is the ratio of the concentrations of the products over the reactants. When a chemical reaction goes to completion, that means that all the reactant has turned into products. As the equilibrium constant defines, it is the ratio of the product to the reactant. So at the final stage of the chemical reaction, the equilibrium constant will be very large.

7 0
4 years ago
Which of the following elements is found in the p sublevel?<br> O Ti<br> O Li<br> ОСІ<br> O He
ICE Princess25 [194]

Answer:

OC I

Explanation:

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8 0
3 years ago
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.
Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0
4 years ago
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