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Eddi Din [679]
2 years ago
9

A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water ________

degrees) Celsius.
Chemistry
1 answer:
Kitty [74]2 years ago
7 0

A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

<h3>What is One Calorie ?</h3>

The amount of heat energy required to raise the temperature by 1 gram of water through 1°C is known as One Calorie.

1 Calorie = 4.18 J

Thus from the above conclusion we can say that A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

Learn more about the One calorie here: brainly.com/question/1061571

#SPJ4

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Answer: 680ml

Explanation:  (450grams)*(0.665g/ml) = 677 ml

6 0
2 years ago
In a chemical reaction, a certain compound changes intoanother compound at a rate proportional to the unchanged amount. ifinitia
Nadya [2.5K]

No it will be a 10% of that balance so take 16 * 10% and then take that answer and divide bye 70%

7 0
3 years ago
1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

6 0
3 years ago
What is the percent by mass of aspartame in iced tea that has 0.75 g of aspartame in 250 g of water?
Vesnalui [34]

Answer:0.30%

Explanation:

6 0
3 years ago
Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t
Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is 1.743\times 10^2g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol

The chemical equation for the hydrogenation of decene follows:

C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = \frac{1}{1}\times 1.225=1.225mol of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g

Hence, the mass of decane produced is 1.743\times 10^2g

5 0
3 years ago
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