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2 years ago

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For lunch, a patient consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.

9). How many kilocalories did the patient obtain from the lunch?

1 answer:

Volgvan2 years ago

3 0

Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk

Energy content of 3 oz skinless chicken is = 110 kcal

Energy content of 3 oz broccoli = 30 kcal

Energy content of 1 medium apple = 60 kcal

Energy content of 1 cup non-fat milk = 90 kcal

So the kilocalories of energy patient obtained from lunch

= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal

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In an ionic bond:

inna [77]

Correct answer is
<span>D. One atom accepts electrons from another.</span>

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2 years ago

Read 2 more answers

Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.

Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × $\frac{254 g}{1 mole}$ = 508 g of Phenobarbital sodium.

1 L × $\frac{1000 mL}{ 1 L}$ = 1000 mL of solution

Thus, % w/v is:

$\frac{508 g}{1000 mL}$ × 100 = 50,8 % w/v

And radio strengths:

$\frac{1000 mL}{508 g}$ = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0

2 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

2 years ago

At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature

Alja [10]

Answer:

Final pressure = 2.3225 atm

Amontons’s law states that

At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature

Explanation:

Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure

Amontons’s law: P/T = Constant at constant V and n

That is P1/T1 = P2/T2

Where temperature is given in Kelvin

Hence T1 of 10°C = 273.15 + 10 = 283.15K

Also temperature T2 of 40°C = 313.15 K

Hence

P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm

3 0

2 years ago

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g × $\frac{1mol}{159,609g}$ = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g × $\frac{1mol}{18,02g}$ = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>

I hope it helps!

5 0

2 years ago