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3 years ago

Lipids that are liquid at room temperature are known as ____.

Chemistry

1 answer:

zaharov [31]3 years ago

7 0

Lipids that are liquid at room temperature are known as oils.

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True or False: A theory is something that can never be proven correct or incorrect?

cestrela7 [59]

Answer:true

true

Explanation:

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2 years ago

Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.

Nookie1986 [14]

100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.

Explanation:

Balanced eaquation for the reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = $\frac{grams of solute}{grams of solution}$

= $\frac{15.2}{57}$

= 0.26 M

Boiling point is calculated as:

ΔT = i x Kb x M

= 1 x 0.512 x 0.26

= 0.133 degrees

The boiling point of the solution will be:

100 degrees + 0.133 degrees (100 degrees is the boiling point of water)

= 100.133 degree celcius is the boiling point of mixture formed.

4 0

3 years ago

Find the enthalpy change per mole of sodium when sodium reacts with water. 13 grams of sodium reacts with 247 cm3 of water, prod

Ierofanga [76]

The answer is

-79 kJ/mol

The explanation:

The reaction equation is:

2Na + 2H2O  2NaOH + H2

when ΔHrxn= -q

and q=cmΔT

where:

-q is amount of heat absorbed/released

-and c is specific heat of solution (in such calculation an assumption is made that c solution = c water)

- and m is mass of solution

- and ΔT is temperature change

1- now we need to Find mass of solution:

m solution = m(H2O)+ m(Na)-m(H2)m(H2O)

m H2O= V(H2O)*ρ(H2O)

= 247 cm3*1g/cm3 = 247 gm

and m(Na)= 13 g

2- now we need to Find m(H2):

n(Na)(moles Na)= mass/Molar mass = 13g/23 g/mol = 0.57 mol.

According to equation mole ratio n(Na):n(H2)= 2:1,

then n(H2)=n(Na)/2

=0.57/2=0.29 mol.

∴m(H2)= moles * molar mass

=0.29 mol *2g/mol

= 0.58 g

∴m solution= 247 g + 13 g – 0.58 g = 259.42 g

and when q=4.18 J/K g* 259.42 g *(339.7 K – 298 K)= 45218 J

The temperature of solution increased because heat was absorbed by the solution (q>0).

Then ΔHrxn= -q = -45218 J per 0.57 mol of Na

3- now we need to Find ΔHrxn per 1 mole of Na

ΔHrxn = -45218J/0.57 mol = -79331 J/mol ≈ -79 kJ/mol

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3 years ago

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Ket [755]

Answer:

Explanation:

statinary front is when a cold air mass and a warm air mass but are at a standstill the boundry is called statinary front sorry i could not give you other anwsers

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2 years ago

What is 140.2 in scientific notation

Akimi4 [234]

Answer:

1.402 * 10^2.

Explanation:

There are 3 digits before the decimal point so the exponent is 2.

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3 years ago

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