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Glucose is a carbohydrate that contains carbon hydrogen and oxygen. The empirical formula of glucose is CH2O and it’s molar mass

shtirl [24]

Glucoses and carbonated

6 0

3 years ago

Consider four different samples: aqueous LiBr , molten LiBr , aqueous AgBr , and molten AgBr . Current run through each sample p

Charra [1.4K]

Answer:

a) Aqueous LiBr = Hydrogen Gas

b) Aqueous AgBr = solid Ag

c) Molten LiBr = solid Li

c) Molten AgBr = Solid Ag

Explanation:

a) Aqueous LiBr

This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.

b) Aqueous AgBr

This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.

c) Molten LiBr

In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.

c) Molten AgBr

The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.

6 0

4 years ago

How many atoms are in 10g of Al

pochemuha

Answer:

The correct answer is

Explanation:

hope this helps u!!

8 0

3 years ago

To what volume would you have to change 6.5 Liters of a gas at 1.25 atm in order to

Salsk061 [2.6K]

Answer:

Explanation:

The required volume can be found by using the formula for Boyle's law

$V_2 = \frac{P_1V_1}{P_2} \\$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

From the question we have

$V_2 = \frac{6.5 \times 1.25}{1} = 6.5 \times 1.25 \\ = 8.125$

We have the final answer as

Hope this helps you

7 0

3 years ago

Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which ar

Marrrta [24]

Answer:

The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

Explanation:

Let the charge on the cation be q₁

Also let the charge on the anion be q₂

A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C

An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C

The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m

Energy of attraction = $\frac{Kq_1q_2}{r^2}$

Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂

Energy of attraction = $\frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}$

Energy of attraction = 1.231 X 10⁻¹¹ J

Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

8 0

3 years ago

SCIENCE PEOPLE HELP ME!​

SCIENCE PEOPLE HELP ME!