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3 years ago

Question 6

Chemistry

2 answers:

Sholpan [36]3 years ago

8 0

Answer:

C 5.0 m/s²

Explanation:

Given data

Mass (m): 9.0 kg

Force (F): 45 N

A 9.0-kg crate rests on a frictionless surface, thus, the only horizontal force to consider is that of 45 N. We can find the acceleration of the 9.0-kg crate due to the 45 N force using Newton's second law of motion.

F = m × a

a = F/m = 45 N/9.0 kg = 5.0 m/s²

Gnom [1K]3 years ago

5 0

Answer:

the answers 5, also did u finish ur ia yet,

if so you know what to do.

Explanation:

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Name three machines which convert straight motion to circular motion.

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A Grinder,A Mixer,A Car

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3 years ago

Is the law of conservative mass observed in this equation CaCO3 + 2HCI -->CaCI2 +H2O + CO2

pychu [463]

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to count how many atoms of each element are there in both sides of the equation:

Ca ⇒ 1 on the left, 1 on the right.

C ⇒ 1 on the left, 1 on the right.

O ⇒ 3 on the left, 3 on the right.

H ⇒ 2 on the left, 2 on the right.

Cl ⇒ 2 on the left, 2 on the right.

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3 years ago

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klemol [59]

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3 years ago

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(01.01 LC)

Usimov [2.4K]

The answer would be "false" or the second option. A opinion is what you think on a certain subject and science is full of facts and opinions of other scientist which would mean your using someone else for your opinion just remember opinions is what you think about a subject and you can't use science for your opinion.

Hope this helps!

7 0

3 years ago

Read 2 more answers

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