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2 years ago

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A preschool has Campbell's Chunky Beef soup, which contains 2.5 g of fat and 15 mg of cholesterol per serving (cup), and Campbel

l's Chunky Sirloin Burger soup, which contains 7 g of fat and 15 mg of cholesterol per serving. By combining the soups, it is possible to get 8 servings of soup that will have 29 g of fat and 120 mg of cholesterol. How many cups of each soup should be used?

1 answer:

Eduardwww [97]2 years ago

5 0

Answer:

6 cups of Campbell's Chunky Beef soup and 2 cups of Campbell's Chunky Sirloin Burger soup should be used.

Explanation:

Let the cups of Campbell's Chunky Beef soup be x

Let the cups of Campbell's Chunky Sirloin Burger soup be y

Total servings or cups required = 8 cups

$x + y=8$ ...(1)

1 cup of Campbell's Chunky Beef soup contains 2.5 g of fat and 15 mg of cholesterol.

Then in x cups of Campbell's Chunky Beef soup have: x(2.5 g+15 mg)

1 cup of Campbell's Chunky Sirloin Burger soup contains 7 g of fat and 15 mg of cholesterol.

Then in y cups of Campbell's Chunky Beef soup have: y(7 g+15 mg)

Total servings or cups must have 29 g of fat and 120 mg of cholesterol.

$x\times (2.5 g+15 mg)+y\times (7 g+15 mg) =29 g+ 120 mg$

1 mg = 0.001 g

$2.515x+7.015y=29.120 g$ ...(2)

On solving equation (1)& (2) we get;

x = 6 , y = 2

6 cups of Campbell's Chunky Beef soup and 2 cups of Campbell's Chunky Sirloin Burger soup should be used.

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It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

tresset_1 [31]

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

$E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J$

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

$E'=\frac{hc}{\lambda }$ (Using planks equation)

$\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}$

$\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm$

$(1 m = 10^9 nm)$

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

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2 years ago

Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p

Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

$-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}$

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2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

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2 years ago

A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51 % fluorine and 49.49 % iron. When th

sergejj [24]

Answer: 0,4278g of F and 0,4191g of Fe

Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.

For Fluorine (F)

0,847g * 0,5051 = 0,4278g of F

For iron (Fe)

0,847 * 0,4949 = 0,4191g of Fe

This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.

At the end, the sum of the elements masses should be the total mass of the compound.

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2 years ago

Read 2 more answers

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago