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nikdorinn [45]
3 years ago
14

This chart shows the global temperature anomaly (the difference of the expected temperature and the actual temperature) over a s

pan of 130 years. Which facts related to climate change does the chart reveal?
Chemistry
2 answers:
polet [3.4K]3 years ago
8 0
I think its 1 3 and 4
AfilCa [17]3 years ago
8 0

Answer:

there is no chart

Explanation:

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When iron rusts in air iron3 oxide is produced. How many moles of oxygen react with 28.0 mol of iron in the rusting reaction 4fe
Marrrta [24]
This the balanced equation based on the question 4Fe + 3O_2  -\ \textgreater \ 2Fe_2O_3.

We then proceed with the following calculations

28.0molFe*(\frac {3molO_2}{4mol Fe})= 21molO_2

The answer is 21molO_2 is produced.
5 0
3 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
3 years ago
Definition of proton and example
gtnhenbr [62]

Answer:

An elementary particle that is identical with the nucleus of the hydrogen atom, that along with the neutron is a constituent of all other atomic nuclei, that carries a positive charge numerically equal to the charge of an electron.

Example:

The nucleus of a hydrogen atom or the H+ ion is an example of a proton. Regardless of the isotope, each atom of hydrogen has 1 proton; each helium atom contains 2 protons; each lithium atom contains 3 protons and so on.

3 0
3 years ago
Write balanced equations for the following reactions
Arte-miy333 [17]

Answer:

See explanation

Explanation:

The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;

238/92U + 4/2 He -------> 241/94Pu + 1/0 n

238/92U + 4/2 He ------> 241/94Pu + 1/0 n

14/7N + 4/2 He------> 17/8O + 1/1 p

56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He

4 0
3 years ago
Given the following unbalanced equation:
antiseptic1488 [7]
<h3>Answer:</h3>

11.84 mol CoF₂

<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

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

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

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                       \displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 11.84 \ mol \ CoF_2
7 0
3 years ago
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