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DochEvi [55]
3 years ago
12

One mole of CO2 is equivalent to

Chemistry
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

44.01 grams

Explanation:

the the mass of 1 mole of CO2 is 44.0 1 grams. there are 6.022×1023 molecules of CO2 in a mole, enough to make the grams in the mole equal the mass.

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Which of these statements best explains why the soil around a volcanic region is fertile?
AVprozaik [17]

Answer:

The correct statement is option c, that is, particles discharged in the air by volcanoes fall to the ground and enrich the soil.  

Explanation:

The eruptions of volcanoes lead to the dispersion of ash over the broader regions surrounding the site of eruption. On the basis of the chemistry of the magma, the ash will be comprising different concentrations of soil nutrients. While the major elements found in the magma are oxygen and silica, the eruptions also lead to the discharging of carbon dioxide, water, hydrogen sulfide, sulfur dioxide, and hydrogen chloride.  

In supplementation, the eruptions also discharge bits of rocks like pyroxene, potolivine, amphibole, feldspar that are in turn enriched with magnesium, iron, and potassium. As an outcome, the areas which comprise huge deposits of the volcanic soil are quite fertile.  

3 0
4 years ago
Read 2 more answers
Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
Students find the pH of Substance A to be 2.1 and Substance B to be 4.4. They are told Substance C is less acidic than Substance
Slav-nsk [51]

Answer:

the range should be 2.2 to 4.3

Explanation:

I think so because the numbers at the left side of the scale from 1 are more acidic so as it increases it's still acidic but lesser so 1 is more acidic than 2 so I used 2.2 as the beginning of the range because it's less acidic than A even though its a greater number and 4.3 is lesser than 4.4 but its still greater on the scale. frankly speaking I don't feel so correct because it's in decimal so try and compare facts thank you

8 0
3 years ago
How are water-based solutions formed?
drek231 [11]

Answer:

D. Solutions are formed when the water’s polar molecules separate the polar molecules of an ionic or molecular compound.

Explanation:

Solutions are homogeneous mixtures formed by interaction between solutes and solvents.

Water based solutions have water as the solvents and mostly ionic and molecular compounds as their solutes.

Water is a polar solvent that is capable of dissolving many compounds by hydrating them. The molecules of water surrounds the solute and forces them  to separate.

6 0
3 years ago
a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter
ladessa [460]

Answer:

The concentration of chloride ion is 2.82\times10^{-3}\;mol/L

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the Cl^- content 100 ppm suggests the presence of 100 mg of Cl^- in 1 L of solution.

The molar mass of Cl^- is equal to the molar mass of Cl atom as the mass of the excess electron in Cl^- is negligible as compared to the mass of Cl atom.

So, the molar mass of Cl^- is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of Cl^- present in 100 mg (0.100 g) of Cl^- is calculated as shown below:

N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol

So, there is 2.82\times10^{-3}\;mol of Cl^- present in 1 L of solution.

5 0
4 years ago
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