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23.07.2020

Carbon dioxide is an example of a molecular compound.
O True
O False

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24.06.2023, solved by verified expert

Bhaskar Singh Bora

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No I think it is false

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Bhaskar Singh Bora

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It’s true because Examples include such familiar substances as water (H2O)
(
H
2
O
)
and carbon dioxide (CO2)
(
CO
2
)
(Figure 3.1.1
3.1.
1
)

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Chemistry

Carbon dioxide is an example of a molecular compound. O True O False

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P Answered by PhD

1

No I think it is false

Chemistry

1. which of the following statements is not true about double-replacement reactions? a. the product may precipitate from solution. b. the product may be a gas. c. the product may be a molecular compound. d. the reactant may be a solid metal. 2. the complete combustion of which of the following substances produces carbon dioxide and water? a. c8h18 b. k2co3 c. cahco3 d. no 3. the equation mg(s) + 2hcl(aq) + h2(g) is an example of which type of reaction? a. combination reaction b. single-replacement reaction c. decomposition reaction d. double-replacement

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P Answered by PhD

The first one is going to likely be A because this is what happens in a precipitation reaction, not in a double replacement reaction. The second one is A because that is Methane, which is a source of fuel. The complete combustion formula is Oxygen + fuel, resulting in CO2 and H2O. The final one is a single-replacement reaction. Hope this helps!

Chemistry

1. which of the following statements is not true about double-replacement reactions? a. the product may precipitate from solution. b. the product may be a gas. c. the product may be a molecular compound. d. the reactant may be a solid metal. 2. the complete combustion of which of the following substances produces carbon dioxide and water? a. c8h18 b. k2co3 c. cahco3 d. no 3. the equation mg(s) + 2hcl(aq) + h2(g) is an example of which type of reaction? a. combination reaction b. single-replacement reaction c. decomposition reaction d. double-replacement

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P Answered by PhD

The first one is going to likely be A because this is what happens in a precipitation reaction, not in a double replacement reaction. The second one is A because that is Methane, which is a source of fuel. The complete combustion formula is Oxygen + fuel, resulting in CO2 and H2O. The final one is a single-replacement reaction. Hope this helps!

Chemistry

To use combustion analysis data to determine an empirical formula a molecular formula expresses the number of each kind of atom in a molecule. for example, the molecular formula for propene, c3h6, indicates three carbon atoms and six hydrogen atoms per molecule. this also means that one mole of propene contains three moles of carbon and six moles of hydrogen. an empirical formula expresses the mole ratio of the elements. the empirical formula for propene is ch2, indicating twice as much hydrogen as carbon. when analyzing unknown compounds in a

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0,1 mol of carbon and 0.2 moles of hydrogen were in the original sample.

Explanation:

The combustion reaction of the compound is

$C_{x} H_{y} O_{z} +O_{2} - CO_{2} + H_{2}O$

And the molar mass of CO2 and H2O are 44g/mol and 18g/mol respectively

All the mass of carbon producing the CO2 must come from the carbon of the unknown compound, thus it is necessary to calculate the moles of carbon in 4.4g of CO2

$\frac{4.4gx12g}{44gx12g/mol} = 0.1 mol$ of Carbon

Also, all the mass of hydrogen producing the H2O must come from the hydrogen of the unknown compound, hence it is necessary to calculate the moles of hydrogen in 1.8g of water

$\frac{1.8gx2g}{18gx1g/mol} = 0.2 mol$ of Hydrogen

Chemistry

To use combustion analysis data to determine an empirical formula a molecular formula expresses the number of each kind of atom in a molecule. for example, the molecular formula for propene, c3h6, indicates three carbon atoms and six hydrogen atoms per molecule. this also means that one mole of propene contains three moles of carbon and six moles of hydrogen. an empirical formula expresses the mole ratio of the elements. the empirical formula for propene is ch2, indicating twice as much hydrogen as carbon. when analyzing unknown compounds in a

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P Answered by Master

0,1 mol of carbon and 0.2 moles of hydrogen were in the original sample.

Explanation:

The combustion reaction of the compound is

$C_{x} H_{y} O_{z} +O_{2} - CO_{2} + H_{2}O$

And the molar mass of CO2 and H2O are 44g/mol and 18g/mol respectively

All the mass of carbon producing the CO2 must come from the carbon of the unknown compound, thus it is necessary to calculate the moles of carbon in 4.4g of CO2

$\frac{4.4gx12g}{44gx12g/mol} = 0.1 mol$ of Carbon

Also, all the mass of hydrogen producing the H2O must come from the hydrogen of the unknown compound, hence it is necessary to calculate the moles of hydrogen in 1.8g of water

$\frac{1.8gx2g}{18gx1g/mol} = 0.2 mol$ of Hydrogen

Chemistry

What elements combined with Strontium, St, in a 1:1 ratio?

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P Answered by PhD

Answer: chalcogens.
Explanation: Strontium is an alkaline earth metal, it always exhibits a degree of oxidation in its compounds +2.
Chalcogens are a group of 6 chemical elements (oxygen O, sulfur S, selenium se, tellurium te, polonium Po) that have an oxidation state of -2 => Chalcogens will combine with strontium in a ratio of 1:1.

Chemistry

LaToya uses 50 newtons (N) of force to pull a 500 N cart. Which statements are correct about calculating LaToya s mechanical advantage? Select all that apply.A. The input force is 50 N.B. The output force is 50 N.C. The input force is 500 N.D. The output force is 450 N.E. The mechanical advantage is 10.F. The mechanical advantage is 100.

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P Answered by Master

Answer:

A

Step-by-step explanation:

The input force is 50 N. But it will not create not any change. No mechanical advantage is observed.

Chemistry

Based on the equation, how many grams of Br2 are required to react completely with 29.2 grams of AlCl3 (5 points)? 2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2

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1

Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry

what is the volume of a container that contains 24.0 grams of N2 gas st 328K and .884 atm?

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P Answered by PhD

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082 Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N ×328 K

Solving:

Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

Chemistry

carbon 14 has a half life of 5730 years how much carbon 14 will remain in a 100 g sample after 11460 years

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P Answered by PhD

Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g

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