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yuradex [85]
2 years ago
10

Write the formula, substitute and solve to the nearest tenth.

Chemistry
1 answer:
Gwar [14]2 years ago
5 0
The answer is that 200 ewwuals ano
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The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni
maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is 2.776*10^{-5} \frac{lb}{gal}

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and  1 mL equals 0.00022 Gallons. So

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}

If 1 μg equals 2.205*10⁻⁹  lb, 2.77 μg how many lb equals?

lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}

You get:

2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}

<u><em>The concentration of this solution in units of pounds per gallon is </em></u>2.776*10^{-5} \frac{lb}{gal}<u><em></em></u>

6 0
3 years ago
Assume the recommended single dose of an antihistamine for a child weighing between 18-23 lbs is 0.75 teaspoons (tsp). if 1.00 t
blagie [28]

We are given that 1 teaspoon is equivalent to 5 mL, therefore 0.75 teaspoon is:

0.75 teaspoon * (5 mL / 1 teaspoon) = 3.75 mL

 

So the mass is density times volume:

mass = (12.5 mg/5 ml) * 3.75 mL

<span>mass = 9.375 mg</span>

7 0
3 years ago
Read 2 more answers
2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?​
Sophie [7]
<h3>Answer:</h3>

\displaystyle 0.5 \ mol \ CO_2

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
  • Compounds

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

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

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                        \displaystyle 0.5 \ mol \ CO_2
7 0
2 years ago
If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?
leva [86]

Answer:

14.434 r.a.m.

Explanation:

  • The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

5 0
3 years ago
calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?
Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as c a^{+} ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to F^{-} ions. This process of acceptance of electrons is termed as reduction.

3 0
3 years ago
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