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larisa86 [58]
3 years ago
9

The colour of RBC is Red .why?​

Chemistry
1 answer:
RoseWind [281]3 years ago
7 0

Answer: fat show the first half and a lot of people in a row in this rr

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What is an element?
s2008m [1.1K]
D -a substance made of more than one type of atom
5 0
3 years ago
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What contribution to atomic theory resulted from albert einstein’s work?
gizmo_the_mogwai [7]

Answer: A new model of the atom that described electrons as being in a cloud

Explanation:

7 0
3 years ago
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Create the Equation: What is the Percent Yield of Ammonia (NH3) if 11.8 g is recovered in a reaction with 7.02 x 10^23 molecules
insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

2

NH

3

(

g

)

tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

5 0
3 years ago
In the first distillation this week, which component of the original solvent mixture makes the larger contribution to the vapor
Olenka [21]

In the first distillation this week, Hexane from the original solvent makes a larger contribution to the vapor pressure of the mixture.

In between hexane and toluene, the hexane will have more vapor pressure contribution in the solution. The boiling point of hexane is much lower than toluene. Therefore, it will evaporate easily at low temperatures and start exerting pressure on the solution.

Hence between hexane and toluene, because of more vapor pressure of hexane and lower boiling point, it will easily evaporate and exerts pressure.

Therefore, from the original solvent, hexane makes a larger contribution to the vapor pressure of the mixture.

To learn more about vapor pressure and hexane, visit: brainly.com/question/28206662

#SPJ4

7 0
1 year ago
What is the mass of 1.58 moles of CH4
HACTEHA [7]
<h3>Answer:</h3>

25.4 g CH₄

<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>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.58 mol CH₄

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

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})
  2. Multiply/Divide:                 \displaystyle 25.359 \ g \ CH_4

<u>Step 4: Check</u>

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

25.359 g CH₄ ≈ 25.4 g CH₄

4 0
3 years ago
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