Balance this equation C2H4 + O2 + CO2 + H2O

A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?

Aliun [14]

<span>In order to solve this problem you must first make sure all your numbers are in like terms. From the density value you can see that it is grams per liter. The first conversion you must do in convert the 125.0 mL value to Liters which you would do by dividing by 1000 because 1 liter is equal to 1000 mL. 125.0 divided by 1000 is 0.125 Liter. Now you will use the density equation to solve. The density equation is density is equal to mass divided by volume. Plug in your known numbers for density and volume. Then solve for mass. So Density (1.269 g/l is equal to mass divided by volume (.125 Liter) You must rearrange the equation to multiple density by volume which is 1.269 times 0.125 which will give you 0.1586. Because the Liters cancel each other out, the answer's unit will be grams. Your final answer is 0.1586 grams.</span>

4 0

2 years ago

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains

Nitella [24]

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of unknown metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

6 0

3 years ago

What is released in the process of an atom moving from an excited level to a lower energy level? (2 points) a photon an electron

Tcecarenko [31]

An electron is the correct answer

3 0

2 years ago

5.00 L of air at 750 mmHg pressure was compressed into a 3.00 L steel cylinder. What is the final pressure? (round to significan

kozerog [31]

Answer:

P2 = 1250mmHg

Explanation:

V1 = 5.0L

P1 = 750mmHg

V2 = 3.0L

P2 = ?

According to Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k/V k = P*V

P1*V1 = P2*V2 = P3*V3 =........=Pn*Vn

P1 *V1 = P2 * V2

Solve for P2

P2 = (P1 * V1) / V2

P2 = (750 * 5.0) / 3.0

P2 = 3750 / 3

P2 = 1250mmHg

The final pressure of the gas is 1250mmHg

5 0

3 years ago

The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.36 ml of water at this temperature?

spin [16.1K]

Mass of 1 ml of water 1 g as density of water 1.00 g/ml at 4° C. So, mass of 2.36 ml of water $\frac{1}{2.36}$ g= 0.423 g.

Molecular mass of water is 18 g which indicates 18 g water contains 6.023 X $10^{23}$ number of water molecules. So, 0.423 g of water contains $\frac{6.023 X 10^{23} X 0.423 }{18} g$ = 0.141 X $10^{23}$ number of water molecules= 1.41 X $10^{22}$ number of water molecules.

5 0

3 years ago

Balance this equation C2H4 + O2 + CO2 + H2O ​

Balance this equation C2H4 + O2 + CO2 + H2O