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miv72 [106K]
2 years ago
8

How many molecules are in 34.7 grams of water?

Chemistry
1 answer:
Anika [276]2 years ago
8 0

Answer:

This part require data such as Avogadro's number and the molar mass of water. But first, let's find the mass of water in the specified volume by making use of the density formula:

Density = mass/volume

1 g/mL = Mass/70 mL

Mass = 70 g

Each water contains 18 grams per mole, and each mole contains 6.022×10²³ molecules of water. Thus,

70 g * 1mole/18 g * 6.022×10²³ molecules/mole = 2.342×10²⁴ molecules of water

Explanation:

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2. A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine:2HI (g) H2(g)
Deffense [45]

Answer:

The value of Kp at this temperature is 7.44*10⁻³

Explanation:

Chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

For the general chemical equation for a homogeneous gas phase system:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients of compounds A, B, C and D, the equilibrium constant Kp is determined by the following expression:

Kp=\frac{P_{C} ^{c} *P_{D} ^{d} }{P_{A} ^{a} *P_{B} ^{b} }

Where Px is the partial pressure of each of the components once equilibrium has been reached and they are expressed in atmospheres. The equilibrium constant Kp depends solely on temperature and is dimensionless.

In the case of the reaction:

2 HI (g) ⇔ H₂ (g) + I₂ (g)

the equilibrium constant Kp is determined by the following expression:

Kp=\frac{P_{H_{2} } *P_{I_{2} } }{P_{HI} ^{2} }

The system comes to equilibrium at 425 °C, and

  • PHI = 0.794 atm
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Replacing:

Kp=\frac{0.0685*0.0685}{0.794^{2} }

Kp=7.44*10⁻³

<u><em>The value of Kp at this temperature is 7.44*10⁻³</em></u>

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I’m struggling with dimensional analysis in chemistry! Will someone please help me with .74 Kcal/min to cal/sec ? Explain and gi
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Answer:

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Explanation:

As we know,

                          1 Kcal  =  1000 cal

So,

                     0.74 Kcal  =  X cal

Solving for X,

                      X =  (0.74 Kcal × 1000 cal) ÷ 1 Kcal

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Also we know that,

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Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,

                                       = 740 cal / 60 sec

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