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3 years ago

Oxygen is a gas with no color or smell a. True b. False

Chemistry

2 answers:

Licemer1 [7]3 years ago

3 0

True it is odorless and it has no color

aleksley [76]3 years ago

3 0

Answer:a.) true

Explanation:

look around you... oxygen is in the air, there is no color or smell.

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A gas sample with a volume of 1,500 cm® is heated from -65 °C to 75 °C. Assuming the

djverab [1.8K]

Answer:

V₂ = 2509.62 cm³

Explanation:

Given data:

Initial volume = 1500 cm³

Initial temperature = -65°C (-65 + 273 = 208 K)

Final temperature = 75°C ( 75 +273 = 348 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 cm³ × 348 K / 208 k

V₂ = 522000 cm³.K / 208 k

V₂ = 2509.62 cm³

7 0

3 years ago

Given that the formula of butane C4H10 the accepted value for the molar mass should be

deff fn [24]

Answer:

carbon mass = 12.01g/mol

hydrogen mass = 1.01g/mol

4 carbon atoms and 10 hydrogen so

12.01 x 4 + 1.01 x 10

48.04g/mol + 10.10g/mol

= 58.14g/mol

4 0

2 years ago

A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit

Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

How do you do these?

Agata [3.3K]

Solve these problems like weighted averages:

The first one:

Multiply the masses (isotope numbers) by the decimal form of the percentage. Add them

0.076 (6) + 0.924 (7) = 6.924

The second one:

0.2 (10) + 0.8 (11) = 10.8

If you think about it, these answers make sense. 6.924 is much closer to 7 than to 6 (since there's a lot more lithium-7 than there is lithium-6). 10.8 is closer to 11 than to 10.

6 0

3 years ago