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2 years ago

If a solution conducts electricity, it is positive evidence that

Chemistry

1 answer:

BlackZzzverrR [31]2 years ago

3 0

Answer:

If a solution conducts electricity, it is positive evidence that solute dissolved in solvent is electrolyte.

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Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−

Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $9.7\times 10^{-6}M^{-1}s^{-1}$

t = time taken = 5.00 days

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

$9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)$

$[A]=0.109M$

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M

8 0

3 years ago

What type of bond would you expect in a compound of oxygen & sodium?

Aleks04 [339]

4Na+O2--›2Na2O
Ionic bond

5 0

3 years ago

80 POINTS!!! ALL PLATO USERS: asap!!!

Crank

This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and

rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course

prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.

Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience

expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness

in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of

academic and professional writing; they will learn to read complex texts with understanding and to write prose of

sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,

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3 years ago

1. How many ATOMS of carbon are present in 7.48 grams of carbon monoxide ?

Salsk061 [2.6K]

Answer:

The answer is 1.61 × 10²³ atoms

Explanation:

To determine number of atoms, we will use the formula below

Number of atoms = number of moles (n) × avogadro's constant (6.02 x 10²³)

n was not provided, hence we will solve for n

n = mass/ molar mass

molar mass of carbon monoxide, CO (where C is 12 and O is 16) is 12 + 16 = 28

mass was provided in the question as 7.48

n = 7.48/28

n = 0.267

Hence,

number of atoms = 0.267 × 6.02 x 10²³

= 1.61 × 10²³ atoms

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3 years ago

When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?

irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

1 mol of NH₄Cl ---------------- 1 mol of AgCl

0.00188 mol of NH₄Cl ------------- x

x = (0.00188 x 1) /1

x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

143.5 grams of AgCl -------------- 1 mol

x -------------------------------------------0.00188 moles of AgCl

x = (0.00188 x 143.5) / 1

x = 0.269 grams of AgCl

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3 years ago