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4 years ago

If 840. mL of linseed oil has a mass of 700. g, calculate the density of linseed oil.

Chemistry

1 answer:

aleksklad [387]4 years ago

6 0

Density can be determined as:

d = m /V

whereas, m is mass of compound

v is volume

d = 700 /840 = 0.83 g /mL

Thus density is 0.83 g /mL

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The elements sodium and chlorine react to form sodium chloride. Does sodium chloride retain any of the characteristic properties

Umnica [9.8K]

Answer:

Explanation:

during chemical change, a new product is formed with entirely different properties with the element.

that's why water does not behave like hydrogen or oxygen but like water.

please mark brainliest

4 0

3 years ago

Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K

Ksp=5.45×10^−13

8 0

3 years ago

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

What is the ratio for calcium bicarbonate

guapka [62]

ca(HCO3)2 this is what i think it is

8 0

3 years ago

BRAINLIESTTTT ASAP!!! PLEASE HELP MEEEE :)

Scilla [17]

Precise is the close proximity of repeated measurements. In order to be precise, you need two or more measurements.

Accurate is the close proximity to the real (or expected) measurement.

For Example:

You are at the grocery store buying watermelons. The sign says that the watermelons all weigh 2 lbs. You weigh 3 of them on the scale next to the watermelon display. Their weights are 1.77, 1.80, and 1.82.

→ The scale is precise because it weighs all 3 of them at nearly the same value.

→ The scale is not accurate because their weights are not close to the expected value of 2.0

You take one of the watermelons to the cashier. The scale at the checkout counter weighs it as 1.99. The scale at the checkout counter is accurate. You cannot determine the precision of the scale at the checkout counter because you have no other values to compare it to.

Answer: Precise CANNOT be determined by one measurement.

Accurate CAN be determined by one measurement.

4 0

4 years ago