Arrange these elements according to increasing reactivity: Br, I, F, Cl

Convert 34.4 lb to g. (1 lb = 453.6 g) A) 7.58 X 10-2 g B) 1.56 X 103 g C) 7.58 X 104 g D) 1.56 X 102 g E) 1.56 X 104 g

Lesechka [4]

Answer:

E) 1.56x10⁴g

Explanation:

Pounds and grams are both units of mass. Pounds are used in UK and USA. Grams is the unit of mass used in the international system of units, SI.

To convert pounds, lb, to grams, g, we need the equivalence between these 2 units of mass:

1lb = 453.6g

That means 1lb weighs 453.6g. 34.4lb weigh:

34.4lb * (453.6g / 1lb) = 1.56x10⁴g. That means right option is:

7 0

3 years ago

Nitrogen monoxide and hydrogen react to form nitrogen and water, like this: 2NO+2 H2(g)-N29)+2H2O(9) Write the pressure equilibr

lara31 [8.8K]

Answer:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp

Explanation:

The reaction is:

2NO + 2H₂ → N₂ + 2H₂O

The expression for Kp (pressure equilibrium constant) would be:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²

There is another expression for Kp, where you work with Kc (equilibrium constant)

Kp = Kc (R.T)^Δn

where R is the Ideal Gases constant

T° is absolute temperature

Δn = moles of gases formed - moles of gases, I had initially

4 0

4 years ago

How many molecules of ethane (C2H6) are present in 0.334 g of C2H6?

zmey [24]

Answer:

6.6253*10²¹molecules of ethane (C₂H₆) are present in 0.334 g of C₂H₆

Explanation:

Avogadro's Number is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.

In this case, being:

C: 12 g/mole
H: 1 g/mole

the molar mass of ethane C₂H₆ is:

C₂H₆: 2*12 g/mole + 6* 1 g/mole= 30 g/mole

Then you can apply the following rule of three: if 30 grams of C₂H₆ are present in 1 mole, 0.334 grams of C₂H₆ in how many moles are present?

$moles of C_{2} H_{6} =\frac{0.334 grams*1 mole}{30 grams}$

moles of C₂H₆=0.011

Finally, taking into account the definition of Avogadro's number, you can apply the following rule of three: if there are 6.023 * 10²³ molecules of C₂H₆ in 1 mole, how many molecules are there in 0.011 moles?

$molecules of C_{2} H_{6} =\frac{0.011 moles*6.023*10^{23}molecules }{1 mole}$