Large particles of hardened lava ejected from a volcano are called

The reactants in the above reaction are hydrogen and nitrogen. The product is ammonia. What is a balanced equation for that

Elanso [62]

N2 + 3H2 (arrow) 2NH3

4 0

3 years ago

Bauxite is the chief ore from which aluminum is extracted. Which of these statements is correct about bauxite?

Eddi Din [679]

Answer : The correct answer is Option D.

Explanation:

Ores are defined as natural occurring substance from which metals and another mineral are extracted.

Bauxite is primary ore of an aluminium metal. it doesn't have definite or specific chemical composition. This is because it contains mixture of hydrous aluminum oxide $Al_2O_3$ , aluminum hydroxides $Al(OH)_3$ , insoluble materials like hematite, quartz ,geothite ,magnetite etc.

From the given options the most appropriate statement which is correct about the bauxite is 'It does not have a definite chemical composition'.

Hence, the correct option Is (D).

6 0

4 years ago

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You are given 230 grams of a substance with a half-life of 0.75 years. How much will remain after 3 years?

Degger [83]

After 3 years, the substance will go through 4 half lives. You can do 230*(1/2)^4, which is 14.375.

6 0

3 years ago

For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a

Volgvan

For this reaction, oxidation number of Carbon in CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced. The reducing agent is the one which provides electrons by oxidizing itself so in this case; CO is the reducing agent while the C in CO oxidized to produce electrons. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

6 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago