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3 years ago

Use the reaction below for the decomposition of sodium azide

Chemistry

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

34.05dm^3 of nitrogen gas

Explanation:

First things first, we need to find the number of moles of Sodium azide. We can do that by using the formula m=nM, mass = no. moles x Molar Mass

Rearrange to solve for no. moles and substituting in the known values and we have:

n = m/M

no. moles = 130.0g / (2x(22.99+3x14.01))

no. moles = 130.0/130.0 (4.s.f.)

no. moles = 1

Now we can use the ratio given in the equation to find the number of moles of Nitrogen that will be made:

1 x 3/2 = 1.5 moles of Nitrogen

Now we use the constant that 1 mole of any gas will always have a volume of 22.7dm^3 at STP.

1.5 x 22.7 = 34.05dm^3 of nitrogen gas.

Hope this helped!

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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

Hi um

nydimaria [60]

It’s( A)Condensation

6 0

3 years ago

Read 2 more answers

N2 + 6e → 2N-3

sveta [45]

Answer:

The correct answer is reduction.

Explanation:

Nitrogen gas reacts with hydrogen gas and get reduced to form ammonia. In this reaction.This is important reaction of atmospheric nitrogen fixation.The reaction is carried out by many nitrogen fixing bacteria such as Azotobacter,Clostridium etc.

N2+3H2+6e- = 2NH3

4 0

3 years ago

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

poizon [28]

We know that to relate solutions of with the factors of molarity and volume, we can use the equation: $M_{1} V_{1} = M_{2} V_{2}$

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M $H_{2} SO_{4}$ is the left side of the equation. Then we have:

$(18 M)(0.050 L)=(4.35M) V_{2}$

We can then solve for $V_{2}$ :

$V_{2}= \frac{(18M)(0.05L)}{4.35M}$ and $V_{2} =0.21 L$ or $210 mL$

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.

7 0

3 years ago

if 5430 J of energy is used to heat 1.25 L of room temp. water (23.0 °C) whats the final temp of the water?

skelet666 [1.2K]

We can use the heat equation,
Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Density = mass / volume

The density of water = 0.997 g/mL

Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL

= 1246.25 g

Specific heat capacity of water = 4.186 J/ g °C.

Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.

4 0

3 years ago