If 30.0 grams of calcium metal react with 18.0 grams of oxygen gas, your calculations show that ________ grams of CaO could be produced from the 30.0 grams of calcium and __________ grams of CaO could be produced from the 18.0 grams of oxygen. The theoretical yield of calcium oxide from this reaction is ________.

Answer:

0.319 M

Explanation:

M1V1 = M2V2

M1 = 0.55M

V1=145 mL = 0.145 L

V2 = 250mL = 0.250L

M2 = M1V1/V2=0.55x0.145/0.250 = 0.319 M

Answer:

Approximately [tex]53.3\; \rm g[/tex].

Explanation:

Lookup Avogadro's Number: [tex]N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}[/tex] (three significant figures.)

Lookup the relative atomic mass of [tex]\rm H[/tex], [tex]\rm S[/tex], and [tex]\rm O[/tex] on a modern periodic table:

(For example, the relative atomic mass of [tex]\rm H[/tex] is [tex]1.008[/tex] means that the mass of one mole of [tex]\rm H\![/tex] atoms would be approximately [tex]1.008\![/tex] grams on average.)

The question counted the number of [tex]\rm H_2SO_4[/tex] molecules without using any unit. Avogadro's Number [tex]N_{\rm A}[/tex] helps convert the unit of that count to moles.

Each mole of [tex]\rm H_2SO_4[/tex] molecules includes exactly [tex](1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}[/tex] of these [tex]\rm H_2SO_4 \![/tex] molecules.

[tex]3.27 \times 10^{23}[/tex] [tex]\rm H_2SO_4[/tex] molecules would correspond to [tex]\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol[/tex] of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of [tex]\rm H_2SO_4[/tex] gives the mass of each mole of [tex]\rm H_2SO_4\![/tex] molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

[tex]\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

Calculate the mass of approximately [tex]0.541389\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:

[tex]\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}[/tex].

(Rounded to three significant figures.)

Answer:

pH = 3.69

Explanation:

HCl is a strong acid. It will disscociates completely into dissoluted ions such that,

[tex][H^+]=2\times 10^{-4}\ M[/tex]

The formula for pH is given by :

[tex]pH=-log[H^+]\\\\pH=-log(2\times 10^{-4})\\\\pH=3.69[/tex]

So, the pH of the solution is 3.69.

The answer to your question is 4.

Answer:

6.2

Explanation:

Answer:

When light is shone on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals. The electrons immediately fall back down to lower energy levels and emit light. This process is responsible for the high luster of metals.

Explanation:

Your well-wisher :-)

Answer:

None

Explanation:

There are two S=O. bonds and two S-O bonds in sulfate ion lewis structure. Sulfur atom is the center atom and four oxygen atoms are located around sulfur atom. There are no lone pairs in the last shell of sulfur atom.

Complete Question:

Give one word for the following.

a. An undesirable constituent of air

b. The constituent of air that helps in burning

c. The thick layer of air surrounding the earth

d. Carbon dioxide does not support this process​

Answer:

a. Pollutant.

b. Oxygen.

c. Atmosphere.

d. Combustion.

Explanation:

a. Pollutant: an undesirable constituent of air. Pollution can be defined as the physical degradation or contamination of the environment through an emission of harmful, poisonous and toxic chemical substances known as pollutants. Some examples of pollutant are carbon monoxide, nitrogen dioxide, ozone, etc.

b. Oxygen: the constituent of air that helps in burning. Oxygen is an element that aids in the combustion (burning) process because it is an oxidizing agent.

c. Atmosphere: the thick layer of air surrounding the earth. It comprises of nitrogen, oxygen, carbon dioxide, argon and other gases in very small proportion.

d. Combustion (burning): Carbon dioxide does not support this process but it is an end product of the process.

Combustion can be defined as an exothermic chemical reaction between physical substances, usually in the presence of oxygen and hydrocarbons to produce heat, light and carbon.

Answer: 2

Explanation:

it’s reflected

Answer:

Transmitted

Explanation:

Because this is when its granted safe passage through the medum

Absorbed: energy goes through and disappears

Reflected: is when light bounces off a surface (mirrors)

Here is my answer. I hope this is helpful.

Answer: If we work it out with direct proportions, the vapor pressure of propanone is 56 degrees Celsius. The atmospheric pressure is about 101.

Explanation: hope this helps.

The vapor pressure of propanone at 50°C is :

- 78° vapor pressure

When the vapor pressure is specifically work through the extents to a temperature rate it response is intensely subordinate on their possess a mass nature.

The vapor weight of propanone is calculated and shape in degree Celsius in rate.

The rate of vapor when it is calculated by 50°C with propanone its at long last gives us a 78° vapor pressure.

Learn more about "Propanone":

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We can use the equation for dilutions that relates concentration and volume:

[tex]M_1V_1=M_2V_2[/tex]

where M is the molarity, V is the volume, and 1 and 2 refer to the initial and final states of the solution, respectively. Here, we are given the molarity of the initial solution and the volumes of the initial and final (diluted) solutions. To find the final concentration (i.e., the molarity of the diluted solution), we would be solving for M₂:

[tex]M_2=\frac{M_1V_1}{V_2} = \frac{(0.55 \text{ M})(145 \text{ mL})}{250 \text{ mL}} \\ M_2 = 0.319 \approx 0.32 \text{ M}.[/tex]

The molarity is given to two significant figures as both our M₁ and V₂ are given to two significant figures.

Note: Although our volumes are in mL instead of L, we do not need to convert them to L for the purposes of our calculation since we would be multiplying our V₁ and V₂ by a common factor that would cancel out in division. All that matters is the ratio between the two volumes, which is the same whether the volumes are in mL or in L.

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We follow the exact same procedure in the second question as we did in the first problem: solve for M₂ given M₁ = 9.02 M, V₁ = 35.0 mL, and V₂ = 45.0 mL:

[tex]M_2=\frac{M_1V_1}{V_2} = \frac{(9.02 \text{ M})(35.0 \text{ mL})}{45.0 \text{ mL}} \\ M_2 = 7.02 \text{ M}.[/tex]

Answer:

0.8338395752301793

Explanation:

I used an online converter (grams to moles for FeCl2)

Answer:

C

Explanation:

The pairing of the cell produced during meiosis and correct example is haploid cell: egg cell (Option C).

In conclusion, the pairing of the cell produced during meiosis and correct example is haploid cell: egg cell (Option C).

Learn more in:

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The given question is incomplete. The complete question is:

Determine how many carbon dioxide molecules are produced if molecules of water are produced

[tex]C_2H_6(g)+O_2 (g)\rightarrow CO2(g)+H_2O(g)[/tex]

Answer: [tex]5.60\times 10^{23}[/tex] molecules of carbon dioxide are produced

Explanation:

The balanced chemical equation is:

[tex]2C_2H_6(g)+7O_2 (g)\rightarrow 4CO2(g)+6H_2O(g)[/tex]

[tex]\text{Moles of water}=\frac{\text{given molecules}}{\text{Avogadros number}}=\frac{8.45\times 10^{23}}{6.023\\times 10^{23}}=1.40moles[/tex]

Accoding to stoichiometry:

6 moles of water are produced along with = 4 moles of carbon dioxide

Thus 1.40 moles of water are produced along with = [tex]\frac{4}{6}\times 1.40=0.93[/tex] moles of carbon dioxide

Molecules of carbon dioxide = [tex]moles\times {\text {Avogadros number}}=0.93\times 6.023\times 10^{23}=5.60\times 10^{23}[/tex]

Answer: combustion and girl

Explanation: C3H6O + 4 O2 → 3 CO2 + 3 H2O COMBUSTION

Answer:

2.59g

Explanation:

Using Avagadro's law equation as follows:

V1/n1 = V2/n2

Where;

V1 = initial volume (litres)

V2 = final volume (litres)

n1 = first amount of gas in grams

n2 = second amount of gas in grams

According to the information provided in this question, v1 = 93.2L, v2 = 10.4L, n1 = 23.2 g, n2 = ?

Using V1/n1 = V2/n2

93.2/23.2 = 10.4/n2

Cross multiply

93.2 × n2 = 23.2 × 10.4

93.2n2 = 241.28

n2 = 241.28/93.2

n2 = 2.588

n2 = 2.59g

The mass of the gas is 2.59g that occupies a volume of 10.4 L under the same conditions.

V₁/n₁ = V₂/n₂

Where

V₁ = initial volume (litres)

V₂ = final volume (litres)

n₁ = first amount of gas in grams

n₂ = second amount of gas in grams

Given:

V₁ = 93.2L,

V₂ = 10.4L,

n₁ = 23.2 g,

To find:

n₂ = ?

On substituting the values:

93.2/23.2 = 10.4/n₂

93.2 * n₂ = 23.2 * 10.4

93.2* n₂ = 241.28

n₂ = 241.28/93.2

n₂ = 2.588

n₂ = 2.59g

Thus, the mass of the gas is 2.59g.

Find more information about Avogadro law here:

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Answer:

Explanation:

Metal elements form positively charged ions called cations because they are located on the left side of the periodic table These elements all have valence electrons in an s orbital. These electrons are relatively easy for the atom to lose to achieve a stable octet of electrons in its outermost energy shell.

Answer:

o

Explanation:

Answer:

Each orbit has a specific energy level.

Explanation:

Bohr model

Explanation:

is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength

Answer:

A. 650 g

Explanation:

The equation of the reaction is given as;

2 HgO(s) --> 2Hg(s)  + O2(l)

From the reaction;

2 mol of HgO produces 1 mol of O2

x mol of HgO would produce 1.50 mol of O2

2 = 1

x = 1.5

Solving for x;

x = 1.5 * 2 / 1 = 3 mol

Converting to mass;

Mass = Number of moles * Molar mass

Mass = 3 mol  * 216.59 g/mol

Mass = 649.77 g ≈ 650. g

Correct option is ;

A. 650 g

Answer:

A substance

Explanation:

Answer:

Describe how a sample of copper chloride crystals could be made from copper carbonate and dilute hydrochloric acid. Add excess copper carbonate to hydrochloric acid in a beaker, stirring until there is no further reaction. Filter the mixture to remove the remaining copper carbonate, then heat the remaining solution to the point of crystallisation.

Explanation: Hope this is right.