On which energy level are the valence electrons for calcium (Ca) ?

A compound sample contains 60.87% c, 4.38% h, and 34.75% o by mass. it has a molar mass of 276.2 g/mol. what are the empirical a

Allushta [10]

Step 1) Assume we have 100 grams of the compound. Thus, we're starting with 60.87 grams C, 4.38 grams H, and 34.75 grams O.

Step 2) Convert the masses of each to moles.

60.87 grams C=5.068276436 mol C

4.38 grams H=4.345238095 mol H

34.75 grams O=2.171875 mol O

step 3) Determine your simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles. (In this case, the smallest number of moles is 2.171875 mol O)

O: $\frac{2.171875}{2.171875} = 1$

H: $\frac{4.345238095}{2.171875} = 2$

C: $\frac{5.068276436}{2.171875} = 2.33$

Step 4) Whenever you're doing an empirical formula problem, and you run into a number that ends with .33, rather than rounding down to a whole number, you need to multiply all your ratios by 3 to obtain whole numbers. Thus, you will get:

O=3

H=6

C=7

Step 5) write the empirical formula using the ratios.

The empirical formula is: C₇H₆O₃

Step 6) The subscripts in the molecular formula of a substance are always whole-number multiples of the subscripts in its empirical formula. This multiple is found by dividing the molecular weight (which was given to us in the problem: 276.2 amu) by the empirical formula weight (C₇H₆O₃ =138.118 amu).

$\frac{276.2}{138.118} = 2$

Step 7) simply multiply the subscripts in the empirical formula by the multiple, 2, and you will get the molecular formula.

The molecular formula is: C₁₄H₁₂O₆

6 0

2 years ago

Which response includes all of the following that are single replacement reactions, and no other reactions?

Zanzabum

Only 2nd reaction is a single replacement reaction

<h3>What is a single replacement reaction?</h3>

A single replacement reaction is a reaction in which an element is replaced by another element in a compound.

It is often called a single displacement reaction

The reactants in this reaction are always pure elements and an aqueous compound.

When this reaction occurs, a new aqueous compound and other pure elements are formed as products.

We can write a general single replacement reaction as:

$AB (aq) + C \rightarrow A + CB(aq)$

Hence, only 2nd reaction is a single replacement reaction that is

$Cl_2(g) + 2NaBr(aq) \rightarrow Br_2(l) + 2NaCl (aq)$

Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: Which response includes all of the following that is single replacement reactions and no other reactions?

$Li_2O(s) + CO_2(g) \rightarrow Li_2CO_3(s)$
$Cl_2(g) + 2NaBr(aq) \rightarrow Br_2(l) + 2NaCl (aq)$
$HNO_3(aq) + KOH (aq) \rightarrow KNO_3(aq) + H_2O (l)$
$SO_3(l) + H_2O \rightarrow H_2SO_4$

Learn more about single replacement reaction:

brainly.com/question/1596829

#SPJ4

4 0

1 year ago

What mass of aluminum has a total nuclear charge of 3.0 C?Aluminum has atomic number 13. Suppose the aluminum is all of the isot

pshichka [43]

Answer:

The value is $m = 6.48*10^{-5} \ g$

Explanation:

From the question we are told that

The total nuclear charge is $Q_t = 3.0 \ C$

The atomic number is $n = 13$

The number of neutrons is $n = 14$

Generally the number of protons is mathematically represented as

$N_p = \frac{Q_t}{p}$

Here p is the charge on a proton with the value $p = 1.60 *10^{-19} \ C$

=> $N_p = \frac{3}{p}$

=> $N_p = \frac{3}{1.60 *10^{-19} }$

=> $N_p = 1.875*10^19 \ protons$

Generally the number of atoms present is mathematically represented as

$N_a = \frac{N_p}{n }$

=> $N_a = \frac{ 1.875*10^19}{13}$

=> $N_a = 1.44*10^{18} \ atoms$

Generally the number of moles of atom present is mathematically represented as

$N = \frac{N_a}{A_n }$

Here $A_n$ is the Avogadro's number with a constant value

$A_n = 6.023*10^{23}$

So

$N = \frac{1.44*10^{18}}{6.023*10^{23} }$

=> $N = 2.4*10^{-6} \ mol$

Generally the total mass is mathematically represented as

$m = N * M$

Here M is the molar mass of aluminum with

$M = 27 \ g/mol$

So

$m = 2.4*10^{-6} * 27$

$m = 6.48*10^{-5} \ g$

8 0

2 years ago

2.a. 3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm-3 copper(II)

Ksenya-84 [330]

Answer:

The limiting reactant is CuSO₄.

Explanation:

The reaction is:

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) (1)

To find the limiting reactant we need to find the number of moles of the reactants.

$\eta_{Fe} = \frac{m}{A_{r}}$

Where:

m: is the mass of iron = 3.26 g

$A_{r}$ : is the standard atomic weight of iron = 55.845 g/mol

$\eta_{Fe} = \frac{3.26 g}{55.845 g/mol} = 0.058 moles$

$\eta_{CuSO_{4}} = M*V$

Where:

M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³

V: is the volume of the solution = 80.0 cm³

First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.

$V = 80.0 cm^{3}*\frac{1 dm^{3}}{(10 cm)^{3}} = 0.080 dm^{3}$

Now, the number of CuSO₄ moles is:

$\eta_{CuSO_{4}} = M*V = 0.200 mol/dm^{3}*0.080 dm^{3} = 0.016 moles$

So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1

$\eta_{Fe} = \frac{1 mol Fe}{1 mol CuSO_{4}}*0.016 moles \: CuSO_{4} = 0.016 moles$

Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.

Therefore, the limiting reactant is CuSO₄.

I hope it helps you!

8 0

2 years ago

2KOH + H2SO4 → K2SO4 + 2H20

mars1129 [50]

19.6 × ( 1 mol KOH / 56 grams KOH )

= 0.35 mol KOH

_________________________________

0.35 mol KOH ×( 2 mol H2O / 2 mol KOH )

= 0.35 mol H20

_________________________________

0.35 mol H2O × ( 18 g H2O / 1 mol H2O )

= 6.3 grams H2O

6 0

2 years ago