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4 years ago

What has more density milk or rubbing alcohol

Chemistry

2 answers:

Korolek [52]4 years ago

7 0

Milk has more dentist then alcohol.

Daniel [21]4 years ago

5 0

Question:

what has more density milk or rubbing alcohol

Answer:

Milk has more density.

Hope this helps!

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Using table 9.4, calculate an approximate enthalpy (in kj) for the reaction of 1.02 g gaseous methanol (ch3oh) in excess molecul

Tanya [424]

Given:

Mass of methanol = 1.02 g

To determine:

Enthalpy for the reaction of 1.02 g of methanol with excess O2

Explanation:

Balanced equation-

2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

The reaction enthalpy is given as:

ΔHrxn = ∑nH°f(products) - ∑nH°f(reactants)

where n = number of moles

H°f = standard enthalpy of formation.

ΔHrxn = [2H°f(CO2(g)) + 4H°f(H2O(g))] - [2H°f(CH3OH(g)) + 3H°f(O2(g))]

= [2(-393.5) + 4(-241.8)]-[2(-201.5) + 3(0)] = -1351.2 kJ

Now, 1 mole of CH3OH = 32 g

The calculated ΔHrxn corresponds to 2 moles of CH3OH. i.e.

The enthalpy change for 64 g of Ch3OH = -1351.2 kJ

Therefore, for 1.02g gaseous methanol we have:

ΔH = 1.02 * -1351.2/64 = -21.5 kJ

Ans: The enthalpy for the given reaction is -21.5 kJ

6 0

3 years ago

Pls help!! will give brainly

____ [38]

Answer:

True

Explanation:

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! LOTS OF POINTS^^

Serggg [28]

1. An isomers, are molecules with the same chemical formula but different structural formula.

2. Isomers of hexane are skeletal isomers. They have the same chemical formula (C₆H₁₄) but have different carbon chain and different chemical properties.
There are five isomers of hexane.

Look at the attachment :)

5 0

3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

4 years ago

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6

weeeeeb [17]

Answer:

Rate of the reaction= 9.92× 10^-5 M² min-1

Explanation:

Using the equation of reaction

2N2O5 ⟶ 4NO2+O2

Rate = k[N2O5]²

From the question k= 6.2×10-4

[N2O5]= 0.4

Rate = 6.2×10-4[0.4]²= 9.92×10-5M² min-1

7 0

3 years ago