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3 years ago

14

When naming a transition metal ion that can have more than one common ionic charage, the numerical value of the chrage is indica

ted by a?

1 answer:

alexandr1967 [171]3 years ago

6 0

The answer is a Roman numeral.
Ex:
1: I
2: II
3: III
4: IV
5: V
6: VI
7: VII
8: VIII
9: IX
10: X

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A reaction mechanism has the following elementary step as the slow step: B2A What is the rate law for this elementary step? O ra

xenn [34]

Answer:

r = k [ B ]

Explanation:

B → 2A

⇒ r = k [ A ]/2 = k [ B ]

3 0

3 years ago

Please help me thanks so much??!!.:;))..I will mark you:))

zubka84 [21]

D for sure hope this helps

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3 years ago

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago

Question 4 to make a 0.500 m solution, one could take 0.500 moles of solute and add 1.00 l of solvent. enough solvent to make 1.

stealth61 [152]

The answer is enough solvent to make 1.00 L of solution. Since molarity is the number of moles of solute in one liter of solution, adding 0.500 mole solute to one liter solvent might not result to a solution with one liter total volume. Less than one liter solvent is first added to dissolve 0.500 mole solute and then the solution is carefully filled with more solvent until the solution reaches to one liter total volume. Hence, the resulting solution is a 0.500M concentration.

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3 years ago

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4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2. mass:

FromTheMoon [43]

The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Moles of hydrochloric acid = 0.385 mol

Mass of chlorine gas =?

Chemical equation:

4HCl + O₂ → 2Cl₂ + 2H₂O

Now we will compare the moles of Cl₂ with HCl.

HCl : Cl₂

4 : 2

0.385 : 2÷4× 0.385 = 0.1925 mol

Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.1925 mol × 71 g/mol

Mass of Cl₂ = 13.6675 g

Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

Learn more about moles here:

brainly.com/question/8455949

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2 years ago