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Margaret [11]
3 years ago
7

blueberry high fiber muffins contain 51% dietary fiber by mass. if a package with a net weight of 12oz contains six muffins, how

many grams of fiber are in each muffin?
Chemistry
2 answers:
pav-90 [236]3 years ago
4 0
If a package with a net weight of 12 oz. contains 6 muffins. how many ... oz. contains 6 muffins. how many grams of fiber are in each muffin?
mars1129 [50]3 years ago
3 0

51% of twelve is 6.12

6.12 divided by 6 is 1.02

your answer is 1.02

You might be interested in
3 elements that have similar properties
qwelly [4]

Answer:The elements in the first column of the Periodic Table (other than hydrogen) are known as Group 1A metals, or alkali metals. When you compare the chemical properties of these elements (lithium, sodium, potassium, rubidium, cesium, and francium), what you'll notice is that they are all remarkably similar.

Explanation:

4 0
3 years ago
Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buff
Alecsey [184]

Answer:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀ \frac{A^{-} }{HA}

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; <em>pka=7,21</em>

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ \frac{HPO4^{2-} }{H2PO4^{-} }

If desire pH is 7,0 you will obtain:

<em>0,617 =  \frac{HPO4^{2-} }{H2PO4^{-} } </em><em>(1)</em>

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] <em>(2)</em>

Replacing (1) in (2) you will obtain:

<em>[H₂PO₄⁻] = 0,0618 M</em>

And with this value:

<em>[HPO₄²⁻] = 0,0382 M</em>

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× \frac{0,0382mol}{1L}× \frac{141,96g}{1mol} = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× \frac{0,0618mol}{1L}× \frac{119,96g}{1mol} = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ \frac{Tris-base }{Tris-H^{+} }

If desire pH is 8,0 you will obtain:

<em>0,841 =  \frac{Tris-base }{TrisH^{+} } </em><em>(3)</em>

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] <em>(4)</em>

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× \frac{0,0457mol}{1L}× \frac{121,1g}{1mol} = 0,553 g of Tris-base

Tris-HCl = 0,1 L× \frac{0,0543mol}{1L}× \frac{157,6g}{1mol} = 0,856 g of Tris-HCl

I hope it helps!

8 0
3 years ago
What rule/principle states that electrons fill orbitals from lowest energy to highest enegery?​
castortr0y [4]

Answer:

The Aufbau Principle

Explanation:

In the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy level before occupying higher-energy levels.

7 0
3 years ago
2 propanoate on oxidation gives ?
TiliK225 [7]

Answer:

l hope it will help u:-)

7 0
3 years ago
Analysis of a compound indicates that it contians 1.04 g K, 0.70 g Cr, and 0.86 g O. Find its empirical formula.
Annette [7]

Answer:

K2CrO4

Explanation:

To find the empirical formula, we need to divide each element by its atomic mass. The atomic masses of potassium, chromium and oxygen are 39.0983, 51.9961 and 15.999 respectively. We make the divisions as follows:

K = 1.04/39.0983 = 0.027

Cr = 0.70/51.9961 = 0.013

O = 0.86/15.999 = 0.054

We now divide by the smallest which is the number of moles of the Chromium

K = 0.027/0.013 = 2

Cr = 0.013/0.013 = 1

K = 0.054/0.013 = 4 approximately

The empirical formula is thus:

K2CrO4

7 0
3 years ago
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