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Chemistry

1 answer:

ozzi3 years ago

3 0

Answer:

For more information about each chemical, view the safety data sheets. Click the Product Resources tab followed by Datasheet. You’ll need these materials: test tube rack test tubes (quantity: 6) test tube labels graduated cylinder wash bottle with distilled water (Tap water is acceptable, but it can skew the experimental results.) pipettes (quantity: 2) forceps 50-milliliter Erlenmeyer ask scoop test tube brush apron goggles gloves pen or re-tip marker small piece of sandpaper chemicals from Edmentum Lab kit: copper metal strip (quantity: 1) iron nails (quantity: 2) zinc metal strips (quantity: 3) copper(II) sulfate solution (10 milliliters) iron(III) nitrate solution (10 milliliters), also called ferric nitrate magnesium sulfate solution (10 milliliters) hydrochloric acid (30 milliliters) sodium bicarbonate (7.5 grams

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Explanation:

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For each pair of salts, determine which of the salts, if any, would be expected to be more soluble in acidic solution than in pu

gogolik [260]

Answer:

AgNO2

Explanation:

The question asks to know which of these two insoluble salts is expected to be more soluble in acidic solution than in pure water.

To answer this question specifically, we need to know if the anions contained in the insoluble salt is a conjugate of a weak acid or that of a weak base.

Generally, the solubility of insoluble salts that contain anions which are conjugates of weak acids increases in the presence of an acidic solution than in water. While, the solubility of insoluble salts that contain anions which are conjugates of strong acids decreases in the presence of an acidic solution.

Having said this, AgNO2 contains NO2 which is the conjugate base of the Trioxonitrate iii acid which is a weak acid. Hence, it is expected to be stronger in acidic solution than in water.

4 0

3 years ago

What is the frequency of a 6.14 x 10^4 m wave

andrew11 [14]

From the information give, I'm going to assume the wave is in a vacuum...

You use the equation: f= $\frac{C}{Wavelength}$ where c is the constant for speed of light ( $3.00 x 10^{8}$

so f= $\frac{3.00x10^{8}}{6.14x10^{4}}$

f= 4.89 x $10^{3}$ Hz (approximately)