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3 years ago

Color, texture, and odor are examples of __________ properties. *

Chemistry

2 answers:

Mrrafil [7]3 years ago

7 0

The answer would be c I think as well

zhenek [66]3 years ago

6 0

Answer:

I think the answer is c)

Explanation:

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Which of the following would be investigated in reaction stoichiometry? Group of answer choices The types of bonds that break an

DENIUS [597]

Answer:

The mass of potassium required to produce a known mass of potassium chloride

Explanation:

Stoichiometry deals with the relationship between amount of substances, mass of substances or volume of substances required in a chemical reaction. Stoichiometric relationships may involve reactants alone or reactants and products. These relationships are normally in the form of simple proportion.

A typical example is our answer option, the mass of potassium required could be used to determine the mass of potassium chloride produced after a balanced reaction equation is written.

7 0

3 years ago

HELP PLEASE.

Dafna11 [192]

This hypothetical process would produce actinium-230.

<h3>Explanation</h3>

An alpha decay reduces the atomic number of a nucleus by two and its mass number by four.

There are two types of beta decay: beta minus β⁻ and beta plus β⁺.

The mass number of a nucleus <em>stays the same</em> in either process. In β⁻ decay, the atomic number <em>increases </em>by one. An electron e⁻ is produced. In β⁺ decay, the atomic number <em>decreases </em>by one. A positron e⁺ is produced. Positrons are antiparticles of electrons.

β⁻ are more common than β⁺ in decays involving uranium. Assuming that the "beta decay" here refers to β⁻ decay.

Gamma decays do not influence the atomic or mass number of a nucleus.

Uranium has an atomic number of 92. 238 is the mass number of this particular isotope. The hypothetical product would have an atomic number of 92 - 2 ⨯ 2 + 1 = 89. Actinium has atomic number 89. As a result, the product is an isotope of actinium. The mass number of this hypothetical isotope would be 238 - 2 ⨯ 4 = 230. Therefore, actinium-230 is produced.

The overall nuclear reaction would involve five different particles. On the reactant side, there is

one uranium-238 atom.

On the product side, there are

one actinium-230 atom,
two alpha particles (a.k.a. helium-4 nuclei),
one electron, and
one gamma particle (a.k.a. photon).

$\;_{\phantom{2}92}^{238} \text{U} \to \;_{\phantom{2}89}^{230} \text{Ac} + \;_{2}^{4} \text{He} + \;_{2}^{4} \text{He} + \text{e}^{-} + \gamma$

Consider: what would be the products if the nucleus undergoes a β⁺ decay instead?

8 0

3 years ago

Subduction zones occur on Earth where dense oceanic crust dives under more buoyant continental crust. These boundaries are chara

Mrrafil [7]

heory of natural selection

8 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

Lubov Fominskaja [6]

This is an incomplete question, here is a complete question.

Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.

Answer: The percent yield of water is, 46.8 %

Explanation : Given,

Mass of $H_2SO_4$ = 72.6 g

Mass of $NaOH$ = 77.0 g

Molar mass of $H_2SO_4$ = 98 g/mol

Molar mass of $NaOH$ = 40 g/mol

First we have to calculate the moles of $H_2SO_4$ and $NaOH$ .

$\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}$

$\text{Moles of }H_2SO_4=\frac{72.6g}{98g/mol}=0.741mol$

and,

$\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}$

$\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $H_2SO_4$ react with 2 mole of $NaOH$

So, 0.741 moles of $H_2SO_4$ react with $0.741\times 2=1.482$ moles of $NaOH$

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $H_2SO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $H_2SO_4$ react to give 2 mole of $H_2O$

So, 0.741 moles of $H_2SO_4$ react to give $0.741\times 2=1.482$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(1.482moles)\times (18g/mole)=26.68g$

Now we have to calculate the percent yield of water.

$\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{12.5g}{26.68g}\times 100=46.8\%$

Thus, the percent yield of water is, 46.8 %

3 0

3 years ago

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g) Complete the following table

Aleksandr [31]

Answer:

1. 0.421 g (C₄H₁₀), 1.51 g (O₂), 1.28 g (CO₂), 0.653 g (H₂O)

2. 4.92 g (C₄H₁₀), 17.6 g (O₂), 14.9 g (CO₂), 7.63 g (H₂O)

3. 6.63 g (C₄H₁₀), 23.7 g (O₂), 20.12 g (CO₂), 10.3 g (H₂O)

4. 7.12 g (C₄H₁₀), 12.1 g (O₂), 10.2 g (CO₂), 8.84 g (H₂O)

5. 252 mg (C₄H₁₀), 903 mg (O₂), 763 mg (CO₂), 390 mg (H₂O)

6. 65 mg (C₄H₁₀), 234 mg (O₂), 198 mg (CO₂), 101 mg (H₂O)

Explanation:

1. First of all, we determine the moles of each reactant.

For the first case:

1.51 g . 1 mol/32 g = 0.0472 moles

Ratio is 13:2, 13 moles of oxygen needs 2 moles of C₄H₁₀ for the combustion,

Therefore 0.0472 mol will react with (0.0472 . 2)/13 = 7.26×10⁻³ mol.

Now we convert the moles to mass:

7.26×10⁻³ mol . 58 g/ 1mol = 0.421 g

Now we use stoichiometry to find the mass of the products.

Ratio is 13:8:10.

13 moles of oxygen can produce 8 moles of CO₂ and 10 moles of water

Then, 0.0472 mol would produce:

(0.0472 . 8)/13 = 0.0290 mol

We convert the moles to mass → 0.0290 mol . 44g /mol = 1.28 g

(0.0472 . 10)/13 = 0.0363 mol

We convert the moles to mass → 0.0363 mol . 18 g /1mol = 0.653 g

2. 4.92 g / 58 g/mol = 0.0848 moles of C₄H₁₀

2 moles of C₄H₁₀ react with 13 moles of O₂

So, 0.0848 moles will react with (0.0848 . 13) / 2 = 0.551 moles

We convert to mass: 0.551 mol . 32 g /mol = 17.6 g

Now we use stoichiometry to find the mass of the products.

Ratio is 13:8:10.

0.551 moles of O₂ will produce:

(0.551 . 8)/13 = 0.339 mol of CO₂

We convert to mass: 0.339 mol . 44g / mol = 14.9 g

(0.551 . 10)/13 = 0.424 mol of H₂O

0.424 mol . 18 g /mol = 7.63 g

3. In this case, we have the mass of one of the product

20.12 g . 1mol / 44 g = 0.457 moles of CO₂

According to stoichiometry:

8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀

Then, 0.457 moles of CO₂ would be produced by:

(0.457 . 13)/ 8 = 0.743 moles of O₂

We convert to mass: 0.743 mol . 32 g/1mol = 23.7 g

(0.457 . 2)/8 = 0.114 moles of C₄H₁₀

We convert to mass: 0.114 mol . 58g/mol = 6.63g

Now we can determine, the mass of produced water:

(0.743 . 10)/13 = 0.571 mol of H₂O . 18g /mol = 10.3 g

4. We convert the moles of water:

8.84 g / 18g/mol = 0.491 moles

According to stoichiometry: 10 moles of water are produced by 13 moles of O₂ and 2 moles of C₄H₁₀

Then 0.491 moles will be produced by:

(0.491 . 10)/ 13 = 0.378 moles of O₂

We convert to mass: 0.378 mol . 32 g/1mol = 12.1 g

(0.491 . 2)/8 = 0.123 moles of C₄H₁₀

We convert to mass: 0.123 mol . 58g/mol = 7.12g

Now we can determine, the mass of produced carbon dioxide:

(0.378 . 8)/13 = 0.232 mol of CO₂ . 44g /mol = 10.2 g

5. Mass of mg, must be converted to grams

252 mg . 1 g/1000 mg = 0.252 g

It is the same as 2.

0.252 g of C₄H₁₀ . 1mol/58 g = 4.34×10⁻³ mol

2 mol of C₄H₁₀ react to 13 moles of O₂ then,

4.34×10⁻³ mol will react with (4.34×10⁻³ mol . 13) / 2 = 0.0282 mol

We convert the grams → 0.0282 mol . 32 g/mol = 0.903 g (903 mg)

0.0282 mol of oxygen will produced:

(0.0282 . 8)/13 = 0.0173 mol of CO₂

We convert to mass: 0.0173 mol . 44g / mol = 0.763 g (763 mg)

(0.0282 . 10)/13 = 0.0217 mol of H₂O

0.0217 mol . 18 g /mol = 0.390 g (390 mg)

6. We define the mass of CO₂ → 198 mg . 1g/1000 mg = 0.198 g

0.198 g / 44g/mol = 4.5×10⁻³ moles of CO₂

According to stoichiometry:

8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀

Then, 4.5×10⁻³ moles of CO₂ would be produced by:

(4.5×10⁻³ . 13)/ 8 = 7.31×10⁻³ moles of O₂

We convert to mass: 7.31×10⁻³ . 32 g/1mol = 0.234 g (234 mg)

(4.5×10⁻³ . 2)/8 = 1.125×10⁻³ moles of C₄H₁₀

We convert to mass: 1.125×10⁻³ mol . 58g/mol = 0.065 g (65 mg)

Now we can determine, the mass of produced water:

(7.31×10⁻³ . 10)/13 = 5.62×10⁻³ mol of H₂O . 18g /mol = 0.101 g (101 mg)

6 0

4 years ago