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4 years ago

What are properties or characteristics shared by Metalloids and Nonmetals?

1 answer:

3 0

Nonmetals have properties opposite those of the metals. The nonmetals are brittle, not malleable or ductile, poor conductors of both heat and electricity, and tend to gain electrons in chemical reactions. Some nonmetals are liquids. These elements are shown in the following figure.

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Calculate the de Broglie wavelength of a shotgun

snow_lady [41]

Answer : The de Broglie wavelength will be $3.68\times 10^{-33}m$

Solution :

The formula used for de Broglie wavelength is:

$\lambda=\frac{h}{mv}$ ..........(1)

where,

$\lambda$ = wavelength = ?

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass = 1.2 g = 0.0012 kg

Conversion used : 1 kg = 1000 g

v = velocity = 150 m/s

Now put all the given values in equation 1, we get:

$\lambda=\frac{6.626\times 10^{-34}Js}{(0.0012kg)\times (150m/s)}$

$\lambda=3.68\times 10^{-33}m$

Therefore, the de Broglie wavelength will be $3.68\times 10^{-33}m$

3 0

3 years ago

How mamy moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess

Anestetic [448]

Answer:

3.62moles

Explanation:

First let us generate a balanced equation for the reaction

2Na + Cl2 —> 2NaCl

Molar Mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl from the balanced equation = 2 x 58.5 = 117g

From the question,

46g of Na produced 117g of NaCl.

Therefore, 83g of Na will produce = (83 x 117)/46 = 211.11g of NaCl

Converting this mass (211.11g of NaCl) to mole, we obtain:

n = Mass /Molar Mass

n = 211.11/ 58.5

3.62moles

8 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

4 years ago

A 150 mL of oxygen gas is collected over water at 20 °C and 758 torr. What volume will the same sample of oxygen occupy at STP w

Stella [2.4K]

Answer:

The volume would be; 136.17 ml

Explanation:

Volume V1 = 150 mL

Temperature T1 = 20°C + 273 = 293 K

Pressure P1 = 758 - 17.54 = 740.46 torr

At STP;

Volume V2 = ?

Pressure P2 = 760 torr

Temperature T2 = 273 K

Using the general gas equation;

P1V1 / T1 = P2V2 / T2

Making V2 subject of formulae;

V2 = P1V1T2 / T1P2

Inserting the values we have;

V2 = 740.46 * 150 * 273 / 293 * 760

V2 = 136.17 ml

3 0

3 years ago

What is the sum of 9260 and 3240

Salsk061 [2.6K]

9260 + 3240= 12,500

If have any questions feel free to ask me!

7 0

3 years ago

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