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xz_007 [3.2K]
3 years ago
6

A drug is supplied as a 40 mg/ml solution in a 50 ml vial. you need to make 10 ml of a 10 mg/ml solution. how much concentrate w

ill be needed? ml how much diluent will be needed? ml
Chemistry
1 answer:
Ymorist [56]3 years ago
8 0

Answer: -

2.5 mL

Explanation: -

Volume of final solution = 10 mL

Strength of final solution = 10 mg / mL

Amount of drug required = Volume of final solution x strength of final solution

= 10 mL x 10 mg/ mL

= 100 mg.

Strength of supplied drug = 40mg/ mL

Volume of supplied drug required = \frac{100 mg}{40 mg/ mL}

= 2.5 mL

Thus to make 10 ml of a 10 mg/ml solution. 2.5 mL will be needed.

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State general trend for metal properties as you go left to right across a period
qwelly [4]

Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.

Electronegativity Trends

Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity. However, the most common scale for quantifying electronegativity is the Pauling scale (Table A2), named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to the qualitative nature of electronegativity. Electronegativity values for each element can be found on certain periodic tables. An example is provided below.


From left to right across a period of elements, electronegativity increases. If the valence shell of an atom is less than half full, it requires less energy to lose an electron than to gain one. Conversely, if the valence shell is more than half full, it is easier to pull an electron into the valence shell than to donate one.

From top to bottom down a group, electronegativity decreases. This is because atomic number increases down a group, and thus there is an increased distance between the valence electrons and nucleus, or a greater atomic radius.

Important exceptions of the above rules include the noble gases, lanthanides, and actinides. The noble gases possess a complete valence shell and do not usually attract electrons. The lanthanides and actinides possess more complicated chemistry that does not generally follow any trends. Therefore, noble gases, lanthanides, and actinides do not have electronegativity values.

As for the transition metals, although they have electronegativity values, there is little variance among them across the period and up and down a group. This is because their metallic properties affect their ability to attract electrons as easily as the other elements.

According to these two general trends, the most electronegative element is fluorine, with 3.98 Pauling units.



6 0
3 years ago
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Goryan [66]
What special structures are needed for green plants?
<span> A.chloroplasts and chlorophyll

</span>
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5 0
2 years ago
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A box has a weight of 120 lbs and the bottom of the box is 12 in2 . What is the pressure the box exerts on the floor?
denis23 [38]
How can I help with this question?
7 0
2 years ago
The valance number of M is 3 . what is the formula of its phosphorus​
lakkis [162]
The formula for phosphorus is P4
3 0
2 years ago
at what temperature in celsius would a gas have volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a p
Alex

Answer:

initial temperature=-3.31^{\circ}C

Explanation:

Assuming that the given follows the ideal gas nature;

P_1=0.723atm

V_1=13.5L

T_1=?

P_2=0.612atm

V_2=17.8L

T_2=28 ^{\circ}C =273+28K=301K

mole of gass will remain same at any emperature:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

putting all the value we get:

T_2 =269.7K =-3.31 ^{\circ} C

initial temperature=-3.31^{\circ}C

6 0
2 years ago
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