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1 year ago

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A 5-ounce cup of raspberry yogurt contains 5.0 g of protein, 3.0 g of fat, and 13.3 g of carbohydrate. The fuel values for prote

in, fat, and carbohydrate are 17, 38, and 17 kJ/g, respectively. What is the fuel value of this yogurt in Calories?

1 answer:

castortr0y [4]1 year ago

6 0

Answer:

as it has 17000 joules and 1 calorie has 4184 joules, this yogurt serving has 4.06 calories

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Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

JulijaS [17]

Answer : The correct option is, $30.9^oC$

Explanation :

Formula used :

$q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

$q$ = heat released = 24 KJ

$m$ = mass of bomb calorimeter = 1.30 Kg

$c$ = specific heat = $3.41J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25.5^oC$

Now put all the given values in the above formula, we get the final temperature of the calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

$24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC$

$T_{final}=30.9^oC$

Therefore, the final temperature of the calorimeter is, $30.9^oC$

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1 year ago

Read 2 more answers

When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac

STatiana [176]

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

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1 year ago

A bag of fertilizer is labeled 10-20-20. What is the actual percentage of phosphorus in the fertilizer?

aleksandr82 [10.1K]

Answer:

20% of phosphorus

Explanation:

A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.

They are often designated NPK fertilizers.

Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.

From the designation,

The actual percentage is 20% of phosphorus.

10% of nitrogen

20% of potassium

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1 year ago

A species has the following MO configuration: (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)2. This substance is:_______.

Goshia [24]

Answer : The correct option is, (a) paramagnetic with two unpaired electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 14 electrons present in the given configuration.

The molecular orbital configuration of molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.

Hence, the correct option is, (a) paramagnetic with two unpaired electrons.

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1 year ago

How much heat is released or absorbed (in kJ) by the system in the reaction of 17.7 g of SF6 with 23.7 g of H2O?

Jlenok [28]

Answer:

+279.744 kJ.

Below is an attachment containing the solution to question.

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