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marshall27 [118]
4 years ago
14

The nutritional information on a cereal box shows that one serving of dry cereal has 90 calories (actually kilocalories). If one

were to burn a serving of cereal, the amount of heat given off would be sufficient to raise the temperature of 1 kg of water in how many degrees Celsius?
Chemistry
1 answer:
umka2103 [35]4 years ago
5 0

Answer:

90 °C

Explanation:

First, we must know the specific heat capacity of water, which is defined as the energy required to heat 1 gram of water by one degree Celsius. The specific heat capacity of water is 1 cal·g⁻¹°C⁻¹.

The equation we will use is Q = mcΔt, where Q is the heat energy, m is the mass, c is the specific heat capacity, and Δt is the temperature change. We will rearrange the equation to solve for Δt and substitute the values:

Δt = Q / (mc) = (90 kcal)(1000 cal/kcal) / (1 kg)(1000 g/kg)(1 cal·g⁻¹°C⁻¹) = 90 °C

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Which temperature is warmer 0° F or 0°C
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Explanation:

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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:

<u>Step 1:</u> 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

<u> Step 2:</u> 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

<u>Step 3:</u> 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

<u>Step 4</u>: 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.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> 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

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3 years ago
Based on the following chemical equation how many hydrogen atoms are present in the products side?
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Answer:

the answer is 6

Explanation:

there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.

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What is a zero order reaction? Give two examples.
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