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spin [16.1K]
3 years ago
3

The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41.0°c and 4

.31 atm is __________ ml.
Chemistry
1 answer:
Ad libitum [116K]3 years ago
0 0
Balance Equation is as followm

                                       F₂  +  CaBr₂   →   CaF₂  +  Br₂

According to equation 1 mole of F₂ reacts with 1 mole of CaBr₂. So,
As,
                           199.89 g CaBr₂ reacts with   =   1 mole of F₂
Then,
                       2.67 g of CaBr₂ will react with   =   X mole of F₂
Solving for X,
                                        X  =  (2.67 g × 1 mole) ÷ 199.89 g

                                        X  = 0.0133 moles of F₂

Now,
Covert moles of F₂ into Volume, Assuming it as Ideal gas,
So, Acc. to Ideal Gas Equation

                                          P V  = n R T
Solving for V,
                                             V  =  n R T / P

Putting Values,

               V  =  (0.0133 mol × 0.0820 atm.L.mol⁻¹.K⁻¹ × 313 K) ÷ 4.31 atm

               V  =  0.0792 L

Converting into mL,
               
                   =  0.0792 × 1000

                   =  79.2 mL

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-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

3 0
3 years ago
Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.
slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0
3 years ago
What chemical property must argon have to make it a suitable choice for light bulbs?
wlad13 [49]
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3 years ago
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Answer:

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rosalindfranklin

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3 0
3 years ago
Determine the change in volume that takes place when a 2.00 L sample of N2 gas is heated from 250 C to 500 C.
Effectus [21]

Answer:

V₂ = 2.96 L

Explanation:

Given data:

Initial volume = 2.00 L

Initial temperature =  250°C

Final volume = ?

Final temperature = 500°C

Solution:

First of all we will convert the temperature into kelvin.

250+273 = 523 k

500+273= 773 k

According to Charles's law,

V∝ T

V = KT

V₁/T₁ = V₂/T₂

V₂ = T₂V₁/T₁

V₂ = 2 L × 773 K / 523 k

V₂ =  1546 L.K / 523 k

V₂ = 2.96 L

4 0
3 years ago
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