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2 years ago

A fruit and oatmeal bar contains 142 nutritional calories. Convert this energy to calories

1 answer:

8 0

Answer is: a fruit and oatmeal bar contains 142000 calories.
A nutritional calorie, or kilocalorie, is equal to 1000 calories.
E = 142 kcal · 1000 cal/kcal.
E = 142 000 cal.
Calorie (cal), or small calorie, is the amount of energy needed to heat one gram of water by one degree Celsius.
One small calorie is approximately 4.2 joules.
A calorie is a unit of energy.

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Notice that " s o 4 " appears in two different places in this chemical equation. s o 2− 4 is a polyatomic ion called "sulfate."

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A "3" should but put in front of
"cas o 4 "

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2 years ago

Name one homogenous mixture from salt, sand and iron

Bezzdna [24]

Answer:

heterogeneous mixture or a chemical mixture

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2 years ago

Which of the following results from acid deposition? View Available Hint(s) Which of the following results from acid deposition?

ASHA 777 [7]

Answer:

Objects made of stone, like marble headstones, are worn away.

Explanation:

Acid deposition on earth is mostly as a result of acid rain. Acid rain forms which oxides of non-metals such as nitrogen and carbon dissolves in rain water to produce weak acid.

This weak acid that is produced easily wears away materials made up of marble and other sculptures. The most active of the acid is weak carbonic acid produced by dissolution of carbon dioxide in rain water.

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2 years ago

What is the boiling point of a solution produced by adding 610 g of cane sugar (molar mass 342.3 g/mol) to 1.4 kg of water? For

wel

Answer:

Boiling point of solution is $100.65^{0}\textrm{C}$

Explanation:

Cane sugar is a non-volatile solute.

According to Raoult's law for a non-volatile solute dissolved in a solution-

$\Delta T_{b}=K_{b}.m$

Where, $\Delta T_{b}$ is elivation in boiling point of solution, $K_{b}$ is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.

Here, $K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}$

610 g of cane sugar = $\frac{610}{342.3}$ moles of cane sugar

= 1.78 moles of cane sugar

So, molality of solution (m) = $\frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}$

Plug in all the values in the above equation, we get-

$\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}$

So, boiling point of solution = $(100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}$

7 0

2 years ago

Consider the following system at equilibrium where H° = 16.1 kJ, and Kc = 6.50×10-3, at 298 K. 2NOBr(g) 2NO(g) + Br2(g)If the VO

Norma-Jean [14]

Answer:

1) The value of Kc : (C.) remains the same

2) The value of Qc : (A.) is greater than Kc

3) The reaction must : (B.) run in the reverse direction to reestablish equilibrium.

4) The number of moles of Br2 will : (B.) decrease

Explanation:

Value of concentration equilibrium constant Kc depends only on temperature. Since temperature remains constant, therefore, Kc remains constant. Decrease in volume means increase in pressure. Increase in pressure favors the side with less gaseous species. Hence, increase in pressure will favor the reverse reaction towards reactants.

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2 years ago