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3 years ago

When iodine is added to foods with no starches it turns

Chemistry

1 answer:

Monica [59]3 years ago

5 0

Answer:

Add Iodine-KI reagent to a solution or directly on a potato or other materials such as bread, crackers, or flour. A blue-black color results if starch is present. If starch amylose is not present, then the color will stay orange or yellow

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What is the correct symbol of an ion containing 17 protons,20 neutrons and 18 electrons

Nataliya [291]

Answer:

The symbol of the ion is A^-

Explanation:

Let A be the symbol of the element.

Proton = 17

Electron = 18

Neutron = 20

Since the element has more electrons than protons, it means it has gain electrons.

From the question given,

The difference between the electron and proton = 18 — 17 = 1

So, the element has gain 1 electron.

Therefore the symbol of the ion is A^-

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3 years ago

Please help me with this question it’s really confusing lol. Question 21. Thank youuuu!! :))

bagirrra123 [75]

Answer:

You must write an article if you think there is a link between it asteroid impact and a dinosaur's extinction. explain the correlation between the two events I'm provide evidence within the article that you write that supports your claims.

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2 years ago

Is fertilizer a mixture compound or element

egoroff_w [7]

I always thought it was a mixture but can also be a compound

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3 years ago

Which list of elements consists of metalloids, only?

rewona [7]

Answer: D

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3 years ago

Read 2 more answers

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

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3 years ago