Ask question

Ask question

All categories

2 years ago

8

Mayonnaise is made up of eggs, oil, and vinegar. Under which category it should be classified? elements compounds solutions coll

oids

2 answers:

krok68 [10]2 years ago

8 0

Mayonnaise is a heterogeneous mix. The correct answer is colloids.

musickatia [10]2 years ago

7 0

<span>Mayonnaise is a heterogeneous mixture. It is not transparent, and its components don't settle after mixing. so there for its a colloids.</span>

You might be interested in

Calculate the PH of a solution 0.030 MH2SO4

Zinaida [17]

Answer:

pH= 2- log3

Explanation:

H2SO4 + H2O -> HSO4^(-) + H30^(+)

0.03M ___ ___

___ 0.03M 0.03M

H30^(+) : C = 0.03M

pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3

4 0

3 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

What size container is needed to hold 7.00 mol of a gas at 342 K and 255 kPa? *

ValentinkaMS [17]

PV = nRT

I only know the valeu of R is you're using atm, so convert kPa to atm.

1 atm = 101.325 kPa

2.52 atm = 255 kPa

2.52atm(x) = 7(.0821)(342)

2.52atm(x) = 196.5

x = 78

78 liters

3 0

2 years ago

If 0.034 moles of potassium nitrate were formed how many moles of lead (II) nitrate did you start

uranmaximum [27]

Answer:

0.017 mole of Pb(NO₃)₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KOH + Pb(NO₃)₂ —> 2KNO₃ + Pb(OH)₂

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produce 2 moles of KNO₃.

Finally, we shall determine the number of mole of Pb(NO₃)₂ required to produce 0.034 mole of KNO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produce 2 moles of KNO₃.

Therefore, Xmol of Pb(NO₃)₂ will react to produce 0.034 mole of KNO₃ i.e

Xmol of Pb(NO₃)₂ = 0.034 / 2

Xmol of Pb(NO₃)₂ = 0.017 mole.

Thus, 0.017 mole of Pb(NO₃)₂ is needed for the reaction.

6 0

3 years ago

Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo

borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

$C_4H_5O_1N_2$

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

$C_8H_{10}O_2N_4$

8 0

1 year ago