All categories

3 years ago

Butter has a specific gravity of 0.86. What is the mass of 2.15 L of butter?

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

7 0

Answer:

The mass of butter is 1849 g.

Explanation:

Given data:

Specific gravity of butter = 0.86

Volume = 2.15 L

Mass = ?

Solution:

Specific gravity of a substance is the ratio of density of substance divided by the density of water.

Thus the density of butter is 0.86 g/mL

Formula:

Conversion from L to mL

2.15 /1000 = 2150 mL

d = m/v

0.86 g/mL = m/2150 mL

m = 1849 g

Thus the mass of butter is 1849 g.

You might be interested in

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

Best regards!

4 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

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

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

How many protons does this atom have?:____

Allisa [31]

Answer:

1.6

2.8

3.6

Explanation:

4 0

3 years ago

Read 2 more answers

describe how the current modern atomic theory and model differs from the model jj Thompson proposed ?

myrzilka [38]

J.J Thompson’s model shows a sphere with electrons that are moving around freely. However, Thompson’s model does not show protons or neutrons. The model that we have today gives a clearer structure showing protons, neutrons, and electrons inside an atom.

6 0

3 years ago

Can y'all give me some examples of physical changes

gogolik [260]

Something that melts, something that changes shapes (for instance, play dough being squished), something that boils, something being mixed or dissolved (but only if it doesn't chemically react), etc. A physical change is a change of state of matter.

4 0

3 years ago