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adell [148]
3 years ago
6

Please helpp!!! due todayyyy Answer the 4 questions

Chemistry
1 answer:
timofeeve [1]3 years ago
5 0
Where are the questions ?
You might be interested in
1. Which statement describes the earliest model of<br><br> the atom?
zloy xaker [14]

Answer:

A). An atom is an indivisible hard-sphere.

Explanation:

The 'Atomic Theory' of Dalton is characterized as the earliest model(came in 1803) which described the atoms as the indivisible and resistant spheres. He <u>used the example of watermelon to elaborate that the atoms of a specific element share similar characteristics</u> and the atoms of distinct elements differ in their mass as well as their size. Thus, <u>option A</u> is the correct answer.

5 0
3 years ago
2Al + 6HCl → 2AlCl3 + 3H2<br> If 85.0 grams of HCl react, how many moles of H2 are produced?
Murrr4er [49]

Answer:

1.17 mol

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles corresponding to 85.0 g of HCl

The molar mass of HCl is 36.46 g/mol.

85.0 g × 1 mol/36.46 g = 2.33 mol

Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl

The molar ratio of HCl to H₂ is 6:3.

2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂

8 0
3 years ago
Find the gram formula mass of each compound.
Doss [256]
The gram formula mass is another term for the mass of one mole of the substance which can be calculated by adding up the masses of the elements comprising the formula unit, molecules, etc.

a. Barium Fluoride (BaF2)
     m = mass of Ba + 2(mass of F)
     m = 137.327 g + 2(19 g)
     m = 175.327 g

b. Strontium Cyanide (Sr(CN)2)
    m  = mass of Sr + 2(mass of C) + 2(mass of N)
    m = 87.62 + 2(12) + 2(14)
    m = 139.62 g

c. Sodium hydrogen Carbonate (NaHCO3)
   m = mass of Na + mass of H + mass of C + 3(mass of O)
   m = 23 + 1 + 12 + 3(16)
    m = 84 g

d. Aluminum sulfide (Al2S3)
   m = 2(mass of Al) + 3(mass of S)
   m = 2(26.98) + 3(32)
   m = 149.96 g
6 0
3 years ago
Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74.0 percent of t
charle [14.2K]

Answer:

Answer:

A)6.6×10^22atoms of Pt in the cube

B)1.4×10^-8m

Explanation:

(a) Calculate the number of Pt atoms in the cube.

an edge length of platinum (Pt) = 1.0 cm.

Then Volume= 1.0 cm×1.0 cm×1.0 cm=1cm^3

Then we have volume of the cube as 1cm^3

Given:

The density Pt = 21.45 g/cm3

the mass of a single Pt atom =3.240 x 10^-22 g

Then with 1atom of the platinum element, we can calculate the number of Pt atoms in the cube as

Density of pt/mass of a single Pt atom

=(21.45 /=3.240 x 10^-22)

=6.6×10^22atoms of Pt in the cube

B)Volume of cube V=4/3πr^3

V= 4/3 ×π×r^3

V= 4.19067r^3

r^3= V/4.19067

But volume is not total volume but just 74% of it, then With 74% of the space inside the cube is taken up by Pt atoms, then we need to find 74% of volume of the cube which is 1cm^3

74/100 ×1= 0.74cm^3

Then our new volume V is 0.74cm^3

r^3=0.74/4.19067×6.620 x 10^22

r^3=2.6674×10^-24

r= 3√2.6674×10^-24

r=1.4×10^-8m

8 0
2 years ago
Problem Page Decide whether these proposed Lewis structures are reasonable. proposed Lewis structure Is the proposed Lewis struc
frutty [35]

The question is incomplete, the complete question is shown in the image attached.

Answer:

1) No, it has the right number of valence electrons but does not satisfy the octet rule.

2) Yes

3) No, it has the wrong number of valence electrons, the correct number is 16

Explanation:

If we look at the first structure in the image(HCN), it easy to see from inspection that nitrogen has three valence electrons as it is normally supposed to have. However, if we count all the electrons around nitrogen, we will notice that they are six instead of eight. Thus nitrogen has not satisfied the octet rule here.

The structure HBr satisfies the octet rule, hence it is a reasonable Lewis structures for the compound shown.

CO2 has 16 valence electrons but the structure shown contains about 20 valence electrons hence it is not a reasonable Lewis structure for the compound.

7 0
4 years ago
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