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Vikentia [17]
3 years ago
10

How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?

Chemistry
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

Grace [21]3 years ago
3 0
Which is an example of using an open-ended question to uncover a problem? O a) "Do you have a problem you'd like addressed today?" b) "Is there a problem? C) "What seems to be the problem?" O d) "Can I help you?"
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A sample of chromium oxide is 76.5hromium by weight. what is the simplest formula of the oxide?
otez555 [7]

CrO and Cr₂O₃ make up the simplest chromium oxide formula.

What name does Cr₂O₃ use?

  • Chromium oxide (Cr₂O₃)sometimes referred to as chromium sesquioxide or chromic oxide, is a compound in which chromium is oxidized to a +3 state. Sodium dichromate is calcined with either carbon or sulfur to produce it.
  • Eskolaite, a mineral that bears the name of the Finnish geologist Pentti Eskola, is a kind of chromium oxide green that may be found in nature. The metallic glassy green surface of this unusual material has an unsettling moss-like look that may be used to conceal oneself in the environment.
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6 0
1 year ago
When you break an iron magnet into two pieces, you get ____.
Rudiy27

Answer:

two north poles and two south poles

Explanation:

A single magnet has a north pole and a south pole. If it is broken into two pieces, then each of the two pieces will have a north pole and a south pole.

No matter how many times or into how many pieces a magnet is broken, the resulting pieces will have two poles each.

5 0
2 years ago
Read 2 more answers
why did mendeleev leave blank spaces on his periodic table? did later discoveries justify his predictions?
schepotkina [342]

Answer:

mendeleev left a space

Explanation:

so the periodic table can be organize

7 0
2 years ago
What are the three states or matter? Whoever answers first, gets a follow bck
Elza [17]
Solid liquid and gas is the answer
8 0
3 years ago
Read 2 more answers
Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.
lina2011 [118]

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!

3 0
3 years ago
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