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Firlakuza [10]
3 years ago
11

For each compound determine whether it is organic or inorganic. C10h16kno9s2 naaso2 hsicl3 (ch)4as2 (bio)2co3 h2p2o7 h2o co2 c6h

12o6
a. Organic
b. Inorganic
Chemistry
1 answer:
Sindrei [870]3 years ago
8 0

a. Organic: C₁₀H₁₆KNO₉S₂; (CH₃)₄As₂; C₆H₁₂O₆

b. Inorganic: NaAsO₂; HSiCl₃; (BiO)₂CO₃; H₂P₂O₇; H₂O; CO₂

Compounds containing <em>both C and H</em> are organic.

Compounds that are <em>not organic</em> are inorganic.

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If 30.0 mL of a 12.0 M hydrochloric acid (HCI) stock solution is diluted to a volume of 500.0 mL,
klemol [59]

Answer:

The concentration of dilute solution is 0.72M

Explanation:

According to the given question

V1=30 ml

 S1 = 12 M

 V2= 500 ml

then S2 =?

  we all know that V1S1=V2S2

                           or  S2= V1S1÷V2

                            or  S2 = 30×12÷500

                             or  S2= 360÷500

                             or  S2  =0.72 M.

The concentration of dilute solution is 0.72 M

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/watch?v=DLzxrzFCyOs

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How many electrons are in the highest occupied energy level of these atoms?
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8 0
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Digital signals maintain their quality over long distances better than analogue signals.

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Explanation:

couldnt make my own so...

5 0
3 years ago
Read 2 more answers
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
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