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3 years ago

How is it possible for one substance to act both polar and nonpolar?

1 answer:

8 0

If the polar bonds are evenly (or symmetrically) distributed, the bond dipoles cancel and do not create a molecular dipole.

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What happens when Ethane is treated with bayers reagent

swat32

Explanation:

यह लेख एक आधार है। जानकारी जोड़कर इसे बढ़ाने में विकिपीडिया की मदद करें।

उस पदार्थ या यौगिक को अभिकर्मक (reagent) कहते हैं जो किसी तंत्र में रासायनिक अभिक्रिया उत्पन्न करने के लिये डाला या मिलाया जाता है। उस पदार्थ को भी अभिकर्मक कहेंगे जिसे यह जांचने के लिये मिलाया जाता है कि कोई अभिक्रिया होती है या नहीं। इस तरह के कुछ वैश्लेषिक अभिकर्मक हैं - फेहलिंग का अभिकर्मक (Fehling's reagent), मिलॉन का अभिकर्मक (Millon's reagent) तथा टॉलीन का अभिकर्मक (Tollens' reagent)।

6 0

3 years ago

The pressure of a sample of helium in a 200. ml. container is 2.0 atm. If the 5 points

just olya [345]

The pressure of the gas = 40 atm

<h3>Further explanation</h3>

Given

200 ml container

P = 2 atm

final volume = 10 ml

Required

Final pressure

Solution

Boyle's Law

At a fixed temperature, the gas volume is inversely proportional to the pressure applied

$\tt \rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

Input the value :

P₂ = P₁V₁/V₂

P₂ = 2 x 200 / 10

P₂ = 40 atm

3 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ

Arisa [49]

Answer: help please !!

Explanation:

4 0

3 years ago

Read 2 more answers

Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0

3 years ago