One quarter of chlorine is chlorine-37 and the rest is Chlorine-35. Calculate the average mass of chlorine

Imagine a drop of water in oil. If you add surfactant, explain how the surfactant molecules organize themselves around the water

Oduvanchick [21]

If you add a surfactant in water-oil mixture, surfactant molecules would form micelles wherein the hydrophobic end of the surfactant faces inside the micelle while the hydrophilic end faces the water molecules. Inside these micelles oil particles are attached.

4 0

3 years ago

How old are the lunar highlands? 1. 1.8 to 2.6 billion years old 2. 3.1 to 3.8 billion years old 3. 4.0 to 4.3 billion years old

Anastaziya [24]

They are about 4.5 billion years old. Hope this helps.

4 0

3 years ago

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How many liters of fat would have to be removed to result in a 5.2 lb weight loss? The density of human fat is 0.94 g/mL?

aksik [14]

I don’t even know I’m just answering for pints

7 0

2 years ago

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac

valentinak56 [21]

Answer : The value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

First we have to calculate the $\Delta G_^o$ .

Formula used :

$\Delta G^o=-RT\times \ln K_p$

Now put all the given values in this formula, we get:

$\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})$

$\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole$

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}$

$Q=8.1\times 10^{5}$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})$

$\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole$

Therefore, the value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

3 0

2 years ago

How do isotopes of a givin element similar

BlackZzzverrR [31]

Answer:

All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.

Explanation:

Hope this <em><u>Helped!</u></em> :D

5 0

3 years ago

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