What is the correcr sum of Cl2+KBr=Br2+KCl

If the mass of a material is 40 grams and the volume of the material is 9 cm^3, what would the density of the material be?

solniwko [45]

Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?

Density is defined as the ratio of mass and volume.

So,

d = m/V

Using the values, we get

d = 40/9 = 4.44 g/cm³

This means the density of material would be 4.44 g/cm³

3 0

3 years ago

What astronomical equiptment did hypatia use?

abruzzese [7]

Hypatia used an astrolabe.

8 0

3 years ago

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

2 years ago

c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a

egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0

3 years ago

On a camping trip, Marta roasts a marshmallow in the campfire. The marshmallow grows, and then turns black.

Mandarinka [93]

Answer:

The answer is D.

Explanation:

4 0

3 years ago

Read 2 more answers