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2 years ago

The diagram shows three isotopes. What element is represented?

Chemistry

2 answers:

Marina CMI [18]2 years ago

5 0

Answer:

a. hydrogen

hope it helps :)

please mark brainliest!!!

lisov135 [29]2 years ago

3 0

Answer:

Where is the diagram?please put the pictu

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Each astronaut produces 8.8 × 102 g CO2 per day that must be removed from the air on the shuttle. If a typical shuttle mission i

suter [353]

The reaction of removing CO2 using LiOH is the following:

2 LiOH + CO2 -----> Li2CO3 + H2O

By solving the amount of CO2 the LiOH can scrub:

(3.50 × 10^4 g LiOH) (1 mol LiOH/ 24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2 it can scrub

<span>Since number of astronaut = 32, 083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>

7 0

3 years ago

Read 2 more answers

How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant

nika2105 [10]

<h3>Answer:</h3>

0.0253 mol H₂O

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

[Given] 0.456 g H₂O (water)

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

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

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.025305 \ mol \ H_2O$

<u>Step 4: Check</u>

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

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0

2 years ago

In the reaction 4Al+3O2→_Al2O3 , what coefficient should be placed in front of the Al2 to balance the reaction?

nika2105 [10]

Sorry I came a lil late,

The answer to your question is, 2.

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p

almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen 40300 g

Carbon 11160 g

Hydrogen 6200 g

Nitrogen 1860 g

Calcium 992 g

Phosphorus 744 g

Other elements 744 g

It is possible to check it adding all the components, the total is 62000 g.

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

Percentage ( %)

Oxygen 65.0

Carbon 18.0

Hydrogen 10.0

Nitrogen 3.0

Calcium 1.6

Phosphorus 1.2

Other elements 1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

Percentage ( %) Decimal number

Oxygen 65.0 65/100 = 0.65

Carbon 18.0 18/100 = 0.18

Hydrogen 10.0 10/100 = 0.10

Nitrogen 3.0 3/100 = 0.03

Calcium 1.6 1.6/100 = 0.016

Phosphorus 1.2 1.2/100 = 0.012

Other elements 1.2 1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

Percentage ( %) Decimal number Mass in grams (g)

Oxygen 65.0 65/100 = 0.65 0.65*62000=40300

Carbon 18.0 18/100 = 0.18 0.18*62000= 11160

Hydrogen 10.0 10/100 = 0.10 0.10*62000= 6200

Nitrogen 3.0 3/100 = 0.03 0.03*62000= 1860

Calcium 1.6 1.6/100 = 0.016 0.016*62000= 992

Phosphorus 1.2 1.2/100 = 0.012 0.012*62000=744

Other elements 1.2 1.2/100 = 0.012 0.012*62000=744

5 0

3 years ago

Use the data in the table above to calculate an approximate ΔH∘ in kilojoules for the synthesis of hydrazine from ammonia:

ankoles [38]

DeltaH formation = deltaH of broken bonds - deltaH of formed bonds

Broken bonds: tiple bond N-N and H-H bond
Formed bonds: N-H and N-N bonds

You also have to take note of the molar coefficients

deltaH formation = <span> [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)]
= (945 + 2*436) - (4*390 + 240)
= 17 kJ/mol

The answer is 17 kJ/mol.</span>

8 0

3 years ago