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3 years ago

Covert 32 ft/sec^2 to m/min^2

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

3 0

Answer:

Explanation:

585.2 meters per minute in 32 feet per second To convert any value in feet per second to meters per minute, just multiply the value in feet per second by the conversion factor 18.288. So, 32 feet per second times 18.288 is equal to 585.2 m/min.

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm

ziro4ka [17]

Answer:

The right answer is "8.81 atm".

Explanation:

Given:

V = 5.00 L

Mass = 4900 g

MW = 32 g/mol

T = 350 K

Now,

Number of moles will be:

$n = \frac{Mass}{MW}$

$=\frac{49.00}{32}$

$=1.53125 \ mole$

By using the ideal gas equation, we get

⇒ $PV=nRT$

or,

⇒ $P=\frac{nRT}{V}$

By substituting the values, we get

$=\frac{1.53125\times 0.0821\times 350}{5.00}$

$=\frac{44.1}{5.00}$

$=8.81 \ atm$

4 0

3 years ago

Is a octopus a carnivore or a herbivore

timofeeve [1]

Carnivore, but some are herbivores

4 0

3 years ago

A sample of nitrogen gas has a volume of 478 cm3 and a pressure of 104.1 kPa. What volume would the gas occupy at 88.2 kPa if th

nadezda [96]

<span>1. </span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 104.1 x 478 / 88.2

6 0

3 years ago

Read 2 more answers

Assume the solubility of benzoic acid in ice-cold water is 1.70 g/L and the solubility of benzoic acid in hot water is 68.0 g/L.

lana [24]

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

4 0

3 years ago