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lukranit [14]
3 years ago
11

You have come down with a nasty cold and would like to understand why the virus that infected you is making you feel so lousy. W

ith a few basic facts, given here within the problem, you can show that each infected cell is now very full of virus particles, resulting in your current condition.
a. Given that a typical cell is about 10 um in diameter, calculate its volume in liters.
b. Find the concentration (in Molarity) of a single virus within that cell.
c. Find the concentration (in Molarity) of 100 virus particles within that cell.
Chemistry
1 answer:
umka21 [38]3 years ago
4 0

Answer:

a) The volume is 5.236x10⁻¹³L

b) The molarity of a single virus is 1.91x10¹² mol/L

c) The molarity for a 100 virus particles is 1.91x10¹⁴ mol/L

Explanation:

a) Given:

D = diameter of the cell = 10 μm

r = radius = 10/2 = 5 μm

The volume of the spherical cell is equal:

V=\frac{4}{3} \pi r^{3} =\frac{4}{3} \pi *(5)^{3} =523.6\mu m^{3}

If 1 μm³ = 1x10⁻¹⁵L, then 523.6 μm³ = 5.236x10⁻¹³L

b) The molarity is:

M=\frac{number-of-moles}{volume-of-solution}

For a single virus within the cell

M=\frac{1}{5.236x10^{-13} } =1.91x10^{12} mol/L

c) For a 100 virus particles the molarity is:

M=\frac{100}{5.236x10^{-13} } =1.91x10^{14} mol/L

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What is the hydrogen ion concentration of a solution with a pH 5? *
nekit [7.7K]

Answer:

In the same way, a solution with a pH of 5 contains 10-5mol/l of hydrogen ions, a solution with a pH of 6 contains 10-6mol/l of hydrogen ions, while the solution with a pH of 7 contains 10-7mol/l of hydrogen ions.

Explanation:

5 0
3 years ago
0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
3 years ago
How many moles of calcium, Ca, are in 5.00 g of calcium?
Rom4ik [11]
<h3>Answer:</h3>

0.125 mol Ca

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

5.00 g Ca

<u>Step 2: Identify Conversions</u>

Molar mass of Ca - 40.08 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 5.00 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca})
  2. Multiply:                             \displaystyle 0.12475 \ mol \ Ca

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.12475 mol Ca ≈ 0.125 mol Ca

3 0
3 years ago
How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
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Answer:

85.34g of NH3

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

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Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

Step 3:

Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

3 0
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