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kirill115 [55]
3 years ago
6

Which material has a crystalline structure at room temperature ( 20 degrees Celsius )

Chemistry
2 answers:
Sauron [17]3 years ago
8 0
It's 2, glass. Water, nitrogen, and sucrose don;t have a crystalline structure.
Musya8 [376]3 years ago
8 0

Answer:

Sucrose. Option 4 is correct.

Explanation:

Crystallinity can be defined as the degree of structural order in a solid.

Let us take a look at each of the material mentioned in the question..

The first one is WATER. Water has the molecular formula of H2O, it is bent at an angle approximately 104.5 degrees. Because of its molecular form as liquid, WATER is amorphous and NOT crystalline.

The second material is GLASS: GLASS do not crystallized when they become hardened.

The third material is Nitrogen, NITROGEN is a non metallic element of group 15. Although solid nitrogen have crystalline modifications, they still exist as amorphous.

The fourth material is sucrose. SUCROSE is a disaccharide. It is a crystalline white solid.

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Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

6 0
2 years ago
A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0
2 years ago
Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el
LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>
8 0
2 years ago
Which of these is a compound?<br> A. Na<br> B. Cl<br> C. NaCl<br> D. C
shtirl [24]

Answer:

Option : C

Explanation:

Hope it helps you!

7 0
2 years ago
Read 2 more answers
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
2 years ago
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