How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7 g Fe? Equation Fe2O2(g) + 3CO(g) -> 3CO2(g) + 2Fe(g)

Answer:

B (immunity) is the answer

Answer:

Is zygote fungi an option??

Answer:

zygote fungi

Explanation:

i got the answer right

Answer:  085.5%

Explanation:

This question is incomplete, the complete question is;

A reaction that proceeds by first-order irreversible kinetics is oxidizing chemical A in a wastewater treatment basin with a mean residence time of 1.5 hours. The reaction rate constant is 2.0 Hr-1.The basin is unbaffled and may be characterized as two completely mixed tanks in series.  If the steady-state influent concentration is 30 mg/l, find the effluent concentration.

If baffles are placed in the basin so that the basin may be characterized as four completely mixed tanks in series, and the mean residence time remains constant, find the effluent concentration.

Answer:

a) (two completely mixed tanks in series) the find the effluent concentration is 4.8 [tex]\frac{mg}{l}[/tex]

b) (four completely mixed tanks in series) find the effluent concentration is 3.2 [tex]\frac{mg}{l}[/tex]

Explanation:

Given the data in the question;

we can determine the effluent concentration of two completely mixed tanks in series for first order irreversible reaction using the following equation;

C = Co ( 1 / ( 1 + K[tex]\frac{t}{n}[/tex] )ⁿ

t is the mean hydraulic residence time for two completely mixed tanks in series ( 1.5 hr)

Co is initial concentration of the influent ( 30 [tex]\frac{mg}{l}[/tex] )

C is final concentration of effluent,

n is the number of tanks series ( 2)

k is rate constant for the given first order reaction( 2[tex]\frac{1}{hour}[/tex] )

so we substitute

C = 30 [tex]\frac{mg}{l}[/tex]  ( 1 / ( 1 + 2[tex]\frac{1}{hour}[/tex] . [tex]\frac{1.5}{2}[/tex] )²

C = 30 [tex]\frac{mg}{l}[/tex]  × ( 1/2.5)²

C = 30 [tex]\frac{mg}{l}[/tex] × 0.16

C = 4.8 [tex]\frac{mg}{l}[/tex]

Therefore, (two completely mixed tanks in series) the find the effluent concentration is 4.8 [tex]\frac{mg}{l}[/tex]

b)

using;

C = Co ( 1 / ( 1 + K[tex]\frac{t}{n}[/tex] )ⁿ

t is the mean hydraulic residence time for two completely mixed tanks in series ( 1.5 hr)

Co is initial concentration of the influent ( 30 [tex]\frac{mg}{l}[/tex] )

C is final concentration of effluent,

n is the number of tanks series ( 4)

k is rate constant for the given first order reaction( 2[tex]\frac{1}{hour}[/tex] )

so we substitute

C = 30 [tex]\frac{mg}{l}[/tex]  ( 1 / ( 1 + 2[tex]\frac{1}{hour}[/tex] . [tex]\frac{1.5}{4}[/tex] )⁴

C = 30 [tex]\frac{mg}{l}[/tex] × ( 1/1.75)²

C = 30 [tex]\frac{mg}{l}[/tex] × 0.107

C = 3.2 [tex]\frac{mg}{l}[/tex]

Therefore, (four completely mixed tanks in series) find the effluent concentration is 3.2 [tex]\frac{mg}{l}[/tex]

Answer: C

Explanation:

It gives kinetic energy to the molecules and it breaks the bonds faster because they jiggle more

Answer:

nice handwrtting

Explanation:

Answer:

yeskkfslszv

Explanation:

frgkkfsssssssss

Answer:

They were relatively flat.

Answer:

71.8%

Explanation:

The percent yield of a process or a chemical reaction is calculated as:

percent yield = actual yield/theoretical yield x 100%

From the problem we have:

theoretical yield = 3.9 kg

actual yield = 2.8 kg

Thus, we calculate the yield of H₂SO₄ production as follows:

percent yield = (2.8 kg/3.9 kg) x 100 = 71.79 % ≅ 71.8%

Answer:

the acceleration of the racecar is 6.5 m/s²

Explanation:

Given;

initial velocity of the racecar, u = 0

final velocity of the racecar, v = 39 m/s

time of motion, t = 6.0 s

The acceleration of the racecar is calculated as;

[tex]a = \frac{v-u}{t} \\\\a = \frac{39-0}{6} \\\\a = \frac{39}{6} \\\\a = 6.5 \ m/s^2[/tex]

Therefore, the acceleration of the racecar is 6.5 m/s²

Answer:

  58.0 g/mol

Explanation:

At STP, one mole occupies 22.4 liters. One mole of the hydrocarbon would have a mass of ...

  (2.59 g/L)(22.4 L/mol) ≈ 58.0 g/mol

Answer:

Answer:

Water is also a product of photosynthesis. This water is produced from the oxygen atoms in the carbon dioxide molecules. The oxygen molecules released into the atmosphere come exclusively from the original water molecules, not from the carbon dioxide molecules...

Hope this helps answer your question!

B r a i n l e i s t ?

Answer:

electric circuits can be dangerous

Explanation:

Answer:

The first one

it limits the amount of electrical energy that can pass through a circuit.

Answer:

Russia es el mejor pais? Vladimir Putin, es el mejor presidente del mundo

Answer: Thermal Equilibrium

Explanation:

Answer:

temperature  

Explanation:

I think that's it i'm sorry if i'm wrong

The one that does not describe a chemical property is bleach is poisonous when swallowed. The correct options are C.

Chemical properties are those properties that show the reactivity of an element. These chemical properties are pH, chemical stability, reactivity, radioactive decay, flammability, toxicity, etc.

Gasoline is a chemical compound, its burning will be a chemical property because it shows how it reacts when it comes in contact with another chemical.

Gold is a pure element, its reactivity tendency is a chemical property, and it is not reactive with oxygen. Density is also a property of an element. And lead is denser than gold, which will show a chemical property.

Thus, the correct options are C. Bleach is poisonous when swallowed.

To learn more about chemical properties, refer to the link:

https://brainly.com/question/5186976

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Answer:

640.32 g

Explanation:

2KClO₃ → 2KCl + 3O₂

First we convert KClO₃ moles to O₂ moles, using the stoichiometric coefficients:

Then we convert O₂ moles to grams, using its molar mass:

So 640.32 g of O₂ are formed from 13.34 moles of KClO₃.

Answer and Explanation:

(a) When lithium bromide (LiBr) solution is mixed with a solution of lead (II) nitrate (Pb(NO₃)₂), lithium nitrate (LiNO₃) and lead (II) bromide (PbBr₂) are formed, according to the following molecular equation:

2LiBr(aq) + Pb(NO₃)₂(aq) → 2LiNO₃(aq) + PbBr₂(s) ↓

As the product PbBr₂ is an insoluble solid, it precipitates (↓).

(b) The total ionic equation is written with all ions of the reaction - no matter if they participate in the precipitate formation or not:

2Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → 2Li⁺(aq) + 2NO₃⁻(aq) + PbBr₂(s)

(c) The net ionic equation is written including only the ions which participate in the precipitate formation. In this case, the precipitate is PbBr₂, and it is formed by Pb²⁺ and Br⁻ ions:

Pb²⁺(aq) + 2Br⁻(aq) → PbBr₂(s)

(d) The spectator ions are those which do not participate in the formation of the precipitate. From the total ionic equation, we can see that Li⁺ and NO₃⁻ ions are repeated on both sides of the equation, so they are redundant. Thus, the spectator ions are Li⁺ and NO₃⁻ ions.

(e) To identify the limiting reactant, we first calculate the moles of each compound as the product of the solution concentration and volume:

For LiBr:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of LiBr = 0.1 L x 1.00 mol/L = 0.1 mol

The same for Pb(NO₃)₂:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of Pb(NO₃)₂  = 0.1 L x 1.00 mol/L = 0.1 mol

From the total ionic equation, we can see that 2 mol of LiBr reacts with 1 mol of Pb(NO₃)₂ to give 1 mol of PbBr₂ (solid product). The stoichiometric molar ratio is 2 mol LiBr/1 mol Pb(NO₃)₂ and we have 0.1 mol of each reactant (0.1 mol LiBr/0.1 mol Pb(NO₃)₂= 1). As 2 mol LiBr/mol Pb(NO₃)₂ > 1 mol LiBr/mol Pb(NO₃)₂, LiBr is the limiting reactant.

(f) From the total ionic equation, we know that 2 moles of LiBr produce 1 mol of PbBr₂. To determine the mass of solid product (PbBr₂) formed, we first multiply the stoichiometric ratio (1 mol PbBr₂/2 mol LiBr) by the actual number of moles of limiting reactant we have (0.1 mol):

moles of PbBr₂ = 0.1 mol LiBr x (1 mol PbBr₂/2 mol LiBr) = 0.05 mol PbBr₂

Finally, we convert the moles of PbBr₂ to gram by using the molar mass of the compound:

Molar mass PbBr₂ = 207.2 g/mol + (2 x 79.9 g/mol) = 367 g/mol

grams of PbBr₂ = 0.05 mol x 367 g/mol = 18.35 g

Explanation:

Structural Formulas v. Empirical Formulas

An empirical formula (like a molecular formula) lacks any structural information about the positioning or bonding of atoms in a molecule. It can therefore describe a number of different structures, or isomers, with varying physical properties. For butane and isobutane, the empirical formula for both molecules is C2H5, and they share the same molecular formula, C4H10. However, one structural representation for butane is CH3CH2CH2CH3, while isobutane can be described using the structural formula (CH3)3CH.

Answer: 1.1 kg

Explanation:

Mw CO2 is 44g —> 6.02214076*10^23 molecules

1.5*10^25 molecules —> 1.1 kg

Answer: Octane will be used completely.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of octane}=\frac{2.3g}{114g/mol}=0.0202moles[/tex]

[tex]\text{Moles of oxygen}=\frac{12.4g}{32g/mol}=0.388moles[/tex]

The balanced chemical reaction will be

[tex]2C_8H_{18}+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]  

According to stoichiometry :

2 moles of octane require = 25 moles of [tex]O_2[/tex]

Thus 0.0202 moles of octane will require=[tex]\frac{25}{2}\times 0.0202=0.2525moles[/tex]  of [tex]O_2[/tex]

Thus octane is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.  

Thus octane will be used completely.

Answer:

0.225 mol = 0.23 mol to 2 significant figures

Explanation:

Calculate the moles of oxygen needed to produce 0.090 mol of water

The equation of the reaction is given as;

2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

From the equation of the reaction;

5 mol of O2 produces 2 mol of H2O

x mol of O2 produces 0.090 mol of H2O

5 = 2

x = 0.090

x = 0.090 * 5 / 2

x = 0.225 mol

Answer:

oxygen, nitrogen, carbon, boron

Explanation:

Answer:

Explanation:

The rocks are composed of mineral grains combined in different ways and having various properties. Minerals are naturally occurring chemical compounds in which atoms are arranged in three-dimensional patterns.

Answer:

7.0

Explanation:

Answer:

pretty sure it's false

Explanation:

Answer:3.50 cm

Explanation:

Answer:

3.63cm

Explanation:

Answer: The smallest structure is the double helix while the largest is the chromosome.

2. G2 phase is the third stage in cellular division wherein the cell prepares itself for the M phase.

3. Interphase comprises of 3 phases, namely, G1 phase, G2 phase, and S phase.

4. Metaphase is known to take less amount of time to complete.

Explanation:

1. The double helix of the DNA or deoxyribonucleic acid, nucleosome, coil, supercoil, and the chromosomes are all linked to the DNA and they vary in their sizes. The nucleosome is known to be the basic unit of the eukaryotic chromosome that is composed of DNA coiled around the histones. The supercoils and coils indicate to the structures which are derived from the folding as well as wrapping of the structure of the DNA. The chromosome refers to the thread-like structure which is made up of nucleic acid and is observed in the nucleus of the living organisms.

2. Cell cycle refers to the cellular division of the cell; it is the series of events which occurs within a cell that results in duplication of the DNA or deoxyribonucleic acid as well as division of the cytoplasm and other organelles in order to generate two daughter cells. This cycle comprises of 4 phases, namely, G1, S, G2, and M phase.

The cell enters the G1 phase wherein the cell synthesizes the proteins as well as mRNA or messenger ribonucleic acid in preparation for the next upcoming phases which lead to mitosis. Next stage is the S phase wherein the synthesis of DNA takes place.

G2 phase is the next phase wherein the cell prepares itself by allowing it to grow as well as generate new proteins needed for the next phase called M phase. Mitosis is the final stage wherein the two identical daughter cells are made from one parent cell.

3. The cell cycle is broadly segregated into 3 phases, namely, interphase, M phase, and cytokinesis. Interphase is the longest stage of the cellular cycle and it comprises of 3 phases, G1 phase, S, and G2 phase. There is a G0 phase or quiescent phase. During the G1 phase, the cell prepares itself for the next phases of cell cycle, S phase involves synthesis of DNA, and G2 is the production of new proteins needed for the M phase.

4. During the mitotic phase, the parental cell segregates and produces 2 identical daughter cells. This phase is further segregated into prophase followed by metaphase, anaphase, and finally telophase. During metaphase, the chromosomes align themselves onto the equatorial or metaphase plate and the centrosomes prepare to segregate the sister chromatids. This phase takes only about 4% of the time needed to finish the entire cell cycle.

Explanation:

Answer:

83

Explanation:

23+60=83