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bixtya [17]
3 years ago
13

Which one of the following substances is a liquid fuel used in rocket engines?

Chemistry
2 answers:
gogolik [260]3 years ago
5 0

Answer:

Condium nitrate

Explanation:

while I do believe it's A if this is for PennFoster I'm almost positive they are looking for C.

uranmaximum [27]3 years ago
3 0
Liquid oxygen can be used as an oxidizer in rocket fuel as one of two propellants needed.
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What is the percent yield for a reaction if we predicted the formation
bogdanovich [222]

Answer:

18.1%

Explanation:

5 0
2 years ago
PLS ANSWER ALL QUESTIONS FOR 67 POINTS
Elena L [17]

Explanation:

1. subatomic particles. 

2.proton, electron and neutron

3.The atomic mass of an element is actually the sum of the MASSES of protons and neutrons in AN atom of that element

4.An element's atomic number is equal to the number of protons in the nuclei of any of its atoms

5. Number of Protons = Atomic Number

Number of Electrons = Number of Protons = Atomic Number

Number of Neutrons = Mass Number - Atomic Number

For krypton:

Number of Protons = Atomic Number = 36

Number of Electrons = Number of Protons = Atomic Number = 36

Number of Neutrons = Mass Number - Atomic Number = 84 - 36 = 48

6. electron, lightest stable subatomic particle known. It carries a negative charge of 1.602176634 × 10−19 coulomb, which is considered the basic unit of electric charge. The rest mass of the electron is 9.1093837015 × 10−31 kg

7.The center of the atom is called a nucleus

8. Negatively charged particles are found in multiple layers outside the nucleus of the atom. These particles are called electrons, and they orbit in various energy levels around the atom's nucleus.

9. A charged particle is also called an ion

3 0
2 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T_2 ),

P_1V_1 / T_1 = P_2V_2 / T_2,

T_2 = P_2V_2T_1 / P_1V_1,

T_2 = 0.73 atm * 1.8 L * 298.15 K / 1 atm * 1.2 L = ( 0.73 * 1.8 * 298.15 / 1 * 1.2 ) K = 326.47425 K,

T_2 = 326.47425 K = 53.32425 C

7 0
3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
Please help me #6!!!!!!!!
Vesnalui [34]
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
3 0
3 years ago
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