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stellarik [79]
3 years ago
10

No one ever helps me ☹️ can y’all look at my questions !?

Chemistry
2 answers:
Orlov [11]3 years ago
8 0

Answer:

   Hope this helps you A typical atom consists of three subatomic particles: protons, neutrons, and electrons (as seen in the helium atom below). Other particles exist as well, such as alpha and beta particles (which are discussed below). The Bohr model shows the three basic subatomic particles in a simple manner. Most of an atom's mass is in the nucleus—a small, dense area at the center of every atom, composed of nucleons. Nucleons include protons and neutrons. All the positive charge of an atom is contained in the nucleus, and originates from the protons. Neutrons are neutrally-charged. Electrons, which are negatively-charged, are located outside of the nucleus. Enjoy your day

Explanation:

Nady [450]3 years ago
4 0

Answer:

Atoms are the fundamental building blocks of all matter and are composed of protons, neutrons, and electrons. Because atoms are electrically neutral, the number of positively charged protons must be equal to the number of negatively charged electrons. Since neutrons do not affect the charge, the number of neutrons is not dependent on the number of protons and will vary even among atoms of the same element.

Explanation:

hope it helps

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how many grams of solid lithium must be added to liquid after in order to obtain 15.0 L of hydrogen gas at STP?
alexdok [17]

Answer:

9.29g

Explanation:

your welcome

3 0
3 years ago
How to calculate from moles to grams?<br> 4.0 moles of Cu(CN)2
andreyandreev [35.5K]

Answer: The mass of given amount of copper (II) cyanide is 462.4 g

Explanation:

To calculate the number of moles, we use the equation:

We are given:

Moles of copper (II) cyanide = 4 moles

Molar mass of copper (II) cyanide = 115.6 g/mol

Putting values in above equation, we get:

Hence, the mass of given amount of copper (II) cyanide is 462.4 g

3 0
3 years ago
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
shutvik [7]

Answer:

  • <u>79%</u>

Explanation:

<u>1) Balanced chemical equation:</u>

  • 2S + 3O₂ → 2SO₃

<u>2) Mole ratio:</u>

  • 2 mol S : 3 mol O₂ : 2 mol SO₃

<u>3) Limiting reactant:</u>

  • Number of moles of O₂

        n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

  • Number of moles of S:

         n = 7.0 g / 32.065 g/mol = 0.2183 mol S

  • Ratios:

        Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

        Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

<u>4) Calcuate theoretical yield (using the limiting reactant):</u>

  • 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

  • x = 0.1875 × 2 / 3 mol SO₃ =  0.125 mol SO₃

<u>5) Yield in grams:</u>

  • mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol =  10.0 g

<u>6) </u><em><u>Percent yield:</u></em>

  • Percent yield, % = (actual yield / theoretical yield) × 100
  • % = (7.9 g / 10.0 g) × 100 = 79%
6 0
3 years ago
A 55.0 g piece of iron at 505 ° C is put into 735 grams of water at 15.0 ° C. What is the final temperature of the water and the
denpristay [2]

Answer:

It would be 95.3

Explanation:

4 0
3 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
BARSIC [14]
<h3>Answer:</h3>

              6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

5 0
3 years ago
Read 2 more answers
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