Ask question

Ask question

All categories

3 years ago

8

What does the kinetic theory describe? A. The transfer of atoms in reactions B. The energy of atomic orbitals C. The reactivity

of valence electrons D. The motion of atoms and molecules

1 answer:

Alenkasestr [34]3 years ago

4 0

I believe the answer you're looking for is, D. The motion of atoms and molecules.

You might be interested in

HELP When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of

babunello [35]

Q=mc*temperature change

400=5.6/1000*c*(87-23)

c=1160.07J/kg/k

4 0

3 years ago

How many moles of FeCl3 could be produced from 6.1 moles of Cly?<br> 2 Fe + 3Cl2 —> 2 FeCl2

Black_prince [1.1K]

Answer:

4.1 moles of FeCl₃

Explanation:

The reaction expression is given as shown below:

2Fe + 3Cl₂ → 2FeCl₃

Number of moles of Cl₂ = 6.1moles

So;

We know that from the balanced reaction expression:

3 moles of Cl₂ will produce 2 moles of FeCl₃

Therefore 6.1moles of Cl₂ will produce $\frac{6.1 x 2}{3}$ = 4.1 moles of FeCl₃

The number of moles is 4.1 moles of FeCl₃

6 0

3 years ago

Does anyone know the answer?????

Juli2301 [7.4K]

Answer:

In chemistry, pH (/piːˈeɪtʃ/) (abbr. power of hydrogen or potential for hydrogen) is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH.

Explanation:

that should answer ur question

3 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

Someone answer the limited regent...

OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

Using Dimensional Analysis

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table

Explanation:

<u>Step 1: Define</u>

[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given] 10 g Cu

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

<u>Step 3: Stoichiometry</u>

<u /> $10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} )$ = 16.974 g Ag

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

16.974 g Ag ≈ 20 g Ag

6 0

3 years ago