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2 years ago

Which starch solution will decrease in volume as osmosis occurs?

1 answer:

iogann1982 [59]2 years ago

7 0

For the following question(s), consider a 4% starch solution and a 10% starch solution separated by a semipermeable membrane.

Which of the following also occurs in this system?
There is a net flow of water from the 4% starch solution into the 10% starch solution

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Compound A, C6H12 reacts with HBr/ROOR to give compound B, C6H13Br. Compound C, C6H14, reacts with bromine and light to produce

Savatey [412]

Answer:

The question involves drawing of structures and showing mechanism in which brainly text editor did not support. I made sure I created a pdf file with both the anwsers and explanations in it. The pdf can be found in the attachment below.

Explanation:

Download pdf

3 0

2 years ago

A scientist compares two samples of white powder. one powder was present at the beginning of an experiment. the other powder was

NARA [144]

Answer. Chemical reaction had occurred and both the powders are different substances.

Explanation:

As density is an intensive property of the substance.Which means that different substance have different densities.

Density = $\frac{mass}{volume}$

Density of powder 1, $d_1=\frac{0.5g}{45cm^3}=0.11g/cm^3$

Density of powder 2, $d_2=\frac{1.3g}{65cm^3}=0.02g/cm^3$

On comparing both the densities of the powders we can say that both the substances are different. So we can conclude that the chemical reaction had occurred.

8 0

2 years ago

Read 2 more answers

How much water(in grams) at its boiling point can be vaporized by adding 1.50 kJ of heat? The molar heat of vaporization for wat

LekaFEV [45]

Answer:

0.66g of water

Explanation:

Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.

If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:

1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.

As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:

0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>

6 0

2 years ago

An irregularly shaped solid which has a mass of 10.283g was placed in a graduated cylinder containing an inert liquid. The initi

Gwar [14]

Liquid + Solid = 8.89 mL
V ( Solid ) = 8.89 mL - 6.26 mL = 2.63 mL
The density of the solid = m / V = 10.283 g / 2.63 mL =
= 3.9 g/mL = 3.9 g / cm³

3 0

2 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

2 years ago