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Brut [27]
3 years ago
10

Maple syrup, which comes from the sap of maple trees, is a mixture of water and natural sugars. It's a clear, brown liquid. It's

homogeneous, and the sugars cannot be separated by filtration or centrifuge. What type of mixture is it?
A. suspension
B. colloid
C. Solution
Chemistry
2 answers:
Margaret [11]3 years ago
5 0
Correct Answer: Option C i.e <span>Solution

Reason: 
Solutions are characterized by particles of size less than 1nm. Since the particle size in solutions are very small, they cannot be separated by centrifugation. On other hand, colloids have particle size ranging from 1nm to 100 nm, while suspensions have particle size > 100 nm. Hence, they can be separated by centrifugation. </span>
jeyben [28]3 years ago
5 0
Maple syrup is a mixture of water and natural sugars. It's a clear, brown liquid that cannot be separated by filtration or centrifuge. It is not a suspension or colloid mixture, it is a solution mixture. The correct answer is C. 
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The document that provides the structure for the us government is
Komok [63]
The Constitution provides the basic structure for US Government.

As a sidenote, you posted this in Chemistry, when it actually belongs in another topic.  Please be sure to post questions only where they belong.  Thanks!  :)
7 0
3 years ago
Develop a demonstration to show how mass is not the same thing as weight
pychu [463]

This is more of a physics explanation, but here we go.

Mass is a measure of how much "matter" is in an object. Weight is the force applied onto an object by gravity. Weight itself can be related to mass like this:

f_g = mg

where g is a gravitational constant. For our purposes, it's defined by whatever planet you are on. Following this, we can demonstrate that mass is NOT the same thing as weight if we take two objects of the same mass and put them on different planets.

Let E refer to Earth and F refer to Mars

g_E = 9.81 m/s^2\\g_F = 3.711 m/s^2

Following this, we can see clearly that weight is not the same as mass:

m*9.81 : m*3.711 \\\int\limits^a_b {x} \, dx 9.81 \neq 3.711\\f_g E \neq f_g F\\

If weight was the same thing as mass, the two values would be the same, as the mass of the two objects is the same. But since weight is defined in the context of gravity, they are not.

4 0
3 years ago
What is the final product of the electron transport chain
Anna71 [15]

Answer:

water and atp

Explanation:

3 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of NH_{3} is as follows:.

          2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)

And, initially only NH_{3} is present.

The given data is as follows.

  P_{NH_{3}} = 2.3 atm at equilibrium

   P_{H_{2}} = 3 \times P_{N_{2}} = 0.69 atm

Therefore,

          P_{N_{2}} = \frac{0.69 atm}{3}

                        = 0.23 aatm

So, P_{NH_{3}} = 2.3 - 2(0.23)

                       = 1.84 atm

Now, expression for K_{p} will be as follows.

         K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}

           K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}

                      = \frac{0.23 \times 0.33}{3.39}

                     = 0.0224

or,           K_{p} = 2.2 \times 10^{-2}

Thus, we can conclude that  the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is 2.2 \times 10^{-2}.

6 0
3 years ago
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