4 Weathering Assessment

What is likely to happen if we can't control our experiment properly? A. Our hypothesis will have to be changed. B. We will need

REY [17]

The answer to your question
is b

4 0

3 years ago

Read 2 more answers

In a sealed gas-liquid system at constant temperature, eventually...?

SOVA2 [1]

<h2>Let us study about it .</h2>

Explanation:

Evaporation

It is the process of converting liquid into vapors .

Condensation

It is the process of converting vapors back into liquid state .

Suppose if we have a sealed container and we are supplying it with no or little heat , we will see that as we increase heat , the particles starts moving faster .
When they move they also colloide and transfer energies .
The kinetic energies of certain molecule increase to an extent that they leave the other particles and escape in atmosphere .
That is evaporation occurs .At the same time when these vapors collide with each other or with the walls of container they get cooled and again get converted to liquid state .

It is seen that a equilibrium is reached when "rate of evaporation becomes equal to rate of condensation ".

4 0

3 years ago

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

Savatey [412]

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0

3 years ago

A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams

kakasveta [241]

Answer:

A. 3.2grams

Explanation:

First we need to get the chemical equation for this reaction:

CO + H₂ → CH₃OH

We then need to balance the equation:

CO + <u>2</u>H₂ → CH₃OH

Our next step is to first convert our given into moles. We do this by figuring out first how many grams of each substance there are in 1 mole by adding up the atomic mass of each element in each substance.

Carbon(1) Oxygen(1)

CO = 12.011(1) + 15.999(1) = 28.01 g/mole

Hydrogen(2)

H₂ = 1.008(2) = 2.016g/mole

We can then use this to determine how many moles of each reactant we have given the mass.

$CO = 2.8g\\2.8g\times\dfrac{1mole}{28.01g}=0.1moles\\\\H_{2}=0.50g\\\\0.50g\times\dfrac{1mole}{2.016g}=0.469moles$

So here we have our new given:

0.1 moles of CO

0.496 moles of H₂

We then need to determine how much product we produce with our given.

According to our chemical equation we can assume that:

For every 1 mole of CO we can produce 1 mole of CH₃OH

Fore every 2 moles of H₂ we can produce 1 mole of CH₃OH

Using this ratio we can determine how much product each reactant will produce by using the ratios:

$0.1 moles of CO\times\dfrac{1moleofCH_{3}OH}{1moleofCO}=0.1moles of CH_{3}OH\\\\ 0.496molesofH_{2}\times\dfrac{1moleofCH_{3}OH}{2molesofH_{2}}=0.248molesofCH_{3}OH$

Now notice that they are not equal. The reactant that produces the least amount of product will be our limiting reactant. The limiting reactant determines the amount of product that is produced, because once it is completely used up, the reaction stops. So in this case, the amount of CH₃OH that is produced is 0.10 moles.

Now we need to figure out how many this is in grams. To do that we need to find out how many grams of CH₃OH there are in 1 mole of CH₃OH.

Carbon (1) Hydrogen(4) Oxygen(1)

CH₃OH = 12.001(1) + 1.008(4) + 15.999(1)

= 12.001 + 4.032 + 15.999 = 32.032g/mole

We then use this for converting:

$0.10 moles\times\dfrac{32.032g}{1mole}=3.2032g$

So the reaction will produce 3.2032g of CH₃OH or 3.2g of CH₃OH

3 0

3 years ago

What is the density of a gas at 242.5K and 0.7311atm. The molar mass of this gas is 70.90g/mol

Pie

Answer:

0.384g/l

Explanation: the density version of the ideal gas law is pm=drt

in which p= pressure, m=molar mass,d=, density, r= to a constant which is 0.08206, and t=temperature so just input the values

PM=DRT. so to find d the formula would be D=RT\PM

D=<u>0.08206*242.5</u>

0.7311*70.90

D=0.384g/l

4 0

3 years ago