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2 years ago

What is the formula, when rubidium reacts with tellurium?

Chemistry

2 answers:

natita [175]2 years ago

5 0

Answer:

Like other alkali metals, rubidium metal reacts violently with water. As with potassium (which is slightly less reactive) and caesium (which is slightly more reactive), this reaction is usually vigorous enough to ignite the hydrogen gas it produces.

Explanation:

hope it helps

kodGreya [7K]2 years ago

4 0

e is the right answer. haha

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Acids, when mixed with water, do which of the following... A Taking away protons from the water molecules B Giving away protons

pav-90 [236]

Answer:

Giving away protons to the water molecules

Explanation:

When acids are mixed with water the following reaction takes place:

H^+(aq) + H2O(l) ------> H3O^+(aq)

Hence when acids are added to water, acids donate a proton to water to form the oxonium ion H3O^+ by coordinate covalent bonding. Note that acids contain the hydrogen ion H^+

4 0

2 years ago

A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two

Galina-37 [17]

Answer:

$16\ \text{ppm}$

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = $2\times 20=40\ \text{mg}$

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = $0.4\times 40=16\ \text{mg}$

So, mass of solute is $16\ \text{mg}$

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is $1\times 1=1\ \text{kg}=1000\ \text{g}$

PPM is given by

$\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}$

Hence, the concentration of iodine in the treated water $16\ \text{ppm}$ .

8 0

2 years ago

6 Which substances have atoms of the same element but different molecular structures?

Bezzdna [24]

Well you want the answer with the same letter in it. A.k.a 2. This means they have the same element in them
O2 (Diatomic oxygen that we breathe) and O3 (Ozone that would be deadly to breathe) have different structures.
So the answer is 2

8 0

2 years ago

Iron fluoride (FeF2) dissociates according to the following equation:

melomori [17]

Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

$Kps = \frac{[product]^x}{[reagent]^y}$

Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.

Analyzing the equation, we see that for 1 mol of $Fe^{+2}$ is necessary 2 mols of $F^-$ , so if we call "x" the molar concentration of $Fe^2$ , for $F^-$ we will have 2x, so:

$Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L$

So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:

Fe = 55.8 g/mol

F = 19 g/mol

FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol

So,

[tex]S = 8.4x10^{-3}x93.8

S = 0.788 g/L

4 0

3 years ago

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

ZanzabumX [31]

The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg.

7 0

2 years ago