What is the product of Pb(NO3)2+KI?

The reaction can be described using the equation: 2C2H25O24CO22H2O. How much C2H2is needed to react with 68.1 g of O2to produce

Sophie [7]

Answer:

22.13g

Explanation:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

2C2H2 + 5O2 —> 4CO2 + 2H2O

Next, we'll calculate the mass of C2H2 and O2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of C2H2 = (12x2) + (2x1)

= 24 + 2 = 26g/mol

Mass of C2H2 that reacted from the balanced equation = 2 x 26 = 52g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 that reacted from the balanced equation = 5 x 32 = 160g

Now, we can obtain the mass of C2H2 that will react with 68.1g of O2 as follow:

From the balanced equation above,

52g of C2H2 reacted with 160g of O2.

Therefore, Xg of C2H2 will react with 68.1g of O2 i.e

Xg of C2H2 = (52x68.1)/160

Xg of C2H2 = 22.13g

Therefore, 22.13g of C2H2 is needed to react with 68.1g of O2

8 0

3 years ago

The reactivity of metals on the periodic table

kari74 [83]

Explanation:

The most reactive metals are found on the left of the periodic table, in the blue column, known as the alkali metals. Their reactivity increases as we go down column (group) one. Reactive metals, when attached to less reactive metals, have the ability to prevent the less reactive metal from rusting.

8 0

3 years ago

An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi

Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow = 713.51 N/m

3 0

2 years ago