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Dovator [93]
3 years ago
11

What Mass of glucose is needed to prepare 235mL of 22% w/v glucose?

Chemistry
2 answers:
rosijanka [135]3 years ago
8 0

Answer: 8 grams of glucose are needed to prepare 400. ml of a 2.0%(m/v) glucose solution. 8 grams of glucose are needed to prepare 400.Oct 6, 2017

uysha [10]3 years ago
4 0

Answer:

8 grams of glucose are needed to prepare 400. ml of a 2.0%(m/v) glucose solution. 8 grams of glucose are needed to prepare 400

Explanation:

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What is the difference between ionic compund and covalent compound​
sleet_krkn [62]

A covalent bond is formed between two non-metals that have similar electronegativities.

An <em>i</em><em>o</em><em>n</em><em>i</em><em>c</em><em> </em><em>b</em><em>o</em><em>n</em><em>d</em> is formed between a metal and a non-metal. Non-metals(-ve ion) are "stronger" than the metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract each other and form the ionic bond.

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4 years ago
H-C C-H name of formula
Vanyuwa [196]

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Ethyne

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7 0
3 years ago
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

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3 years ago
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C because burning waste does not affect the marine <span>ecosystems</span>
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Alexa is teaching students to measure and compare the speed of a pencil as it is dropped from different heights
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