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Nastasia [14]
2 years ago
5

Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a __________

__ mixture. A) heterogeneous B) heterologous C) homogeneous D) homologous
Chemistry
2 answers:
Vesna [10]2 years ago
7 0
Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous
frosja888 [35]2 years ago
6 0

Answer is: A) heterogeneous.

This sample of sand is heterogeneous mixture.

Heterogeneous mixture is not uniform in composition (in this mixture different particles and minerals), but proportions of its components (in this mixture particles of different colors, small pebbles) vary throughout the sample .

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HELP
agasfer [191]

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02  mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02  mol ×  6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

5 0
1 year ago
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
1 year ago
Read 2 more answers
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
2 years ago
Consider the reaction between nis2 and o2: 2nis2(s)+5o2(g)→2nio(s)+4so2(g) when 11.2 g of nis2 are allowed to react with 5.43 g
stellarik [79]
M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.


6 0
1 year ago
Products or solutions found at home or in store and their characteristics
AleksAgata [21]

Answer:

bruh, just go in your bathroom and look for cleaning products

Explanation:

4 0
1 year ago
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