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natka813 [3]
3 years ago
6

Which of the following do water, rust, and salt have in common?

Chemistry
2 answers:
Cloud [144]3 years ago
6 0
The answer is "elements" :)
professor190 [17]3 years ago
3 0

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Atoms are defined as the smallest unit which cannot be further divided.

Mixtures are defined as the solution win which components are mixed in indefinite proportions. Every substance retain their individual chemical properties.

Elements are defined as the substances which are made from only one type of atoms. Example: Sodium (Na), Oxygen (O), etc...

Compounds are defined as the substances which are made from two or more than two types of atoms in a fixed ratio by their mass. For Example:  H_2O,CO_2 etc..

We are given:

Three substances having chemical formulas of H_2O,Fe_2O_3\text{ and }NaCl

All the substances are made from two types of atoms. Thus, they are all considered as compounds.

Hence, the correct answer is Option D.

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vodka [1.7K]

Answer:

C.

Explanation:

We get it from the food we eat by mixing it with fluids. Also, you can always look up the answer.

Hope this helps! :) Plz mark as brainliest!

7 0
3 years ago
Read 2 more answers
8.38e-21 Q^2 -1.07e-23 Q^2 +3.15e+19<br> Solve for Q
GarryVolchara [31]
8.38e -21Q^2 -1.07e -23Q^2 +3.15e +19
= 10.46e -44Q^2 + 19
44Q^2 = 10.46e + 19
Q^2 = 523/2200e + 19/44
Q1= ≈ -1.03828
Q2= ≈ 1.03828
4 0
4 years ago
An unknown substance undergoes a chemical change that gives off heat. What can you tell me about this substance?
marin [14]
On a molecular level there is a lot of movement which in turn is the reason why heat is generated.
7 0
3 years ago
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Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
3 years ago
Copper crystallizes in a face-centered cubic unit cell. the density of copper is 8.94 g/cm3. calculate the length of the edge of
Ilia_Sergeevich [38]
<span>361.4 pm is the length of the edge of the unit cell. First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight. Atomic weight copper = 63.546 Now divide the mass by the atomic weight, getting 8.94 g / 63.546 g/mol = 0.140685488 mol And multiply by Avogadro's number to get the number of atoms: 0.140685488 * 6.022140857x10^23 = 8.472278233x10^22 Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume. 8.472278233x10^22 / 4 = 2.118069558x10^22 Now calculate the volume each unit cell occupies. 1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3 Let's get the cube root to get the length of an edge. (4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm Now let's convert from cm to pm. 3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
4 0
3 years ago
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