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2 years ago

Which set is an example of an element, a compound, and a mixture?

Chemistry

1 answer:

Nitella [24]2 years ago

4 0

Answer:

Element: Sulfur ; S

Compound: Water ; H2O

Mixture: Gunpowder ; 2 KNO3 + S + 3 C -> K2S + N2 + 3 CO2

Explanation:

An element is a pure substance that cannot be broken down into simpler substances. A periodic table for example has elements.

A compound is a chemical substance that's made of multiple elements held together by chemical bonds. All elements from the periodic table can be combined to make a compound but they have to bond a specific way in order for them to actually make something. Water is the easiest to remember.

A mixture is two or more substances that are physically combined. They are not chemically combined though. There are a few types of mixtures including homogeneous, heterogeneous, etc. Gunpowder is actually a mixture. It contains sulfur, carbon and potassium nitrate.

Don't be afraid to reach out with further questions, I hope this helps!

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Enter a balanced equation for the reaction between solid nickel(II)(II) oxide and carbon monoxide gas that produces solid nickel

velikii [3]

Answer: A balanced equation for the given reaction is $NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$ .

Explanation:

The reaction equation will be as follows.

$NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$

Number of atoms on the reactant side is as follows.

O = 2
C = 1

Number of atoms on the product side is as follows.

Ni = 1
O = 2
C = 1

Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.

Thus, we can conclude that a balanced equation for the given reaction is $NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)$ .

7 0

3 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

2 years ago

What is made when a salt is dissolved in water?

sergey [27]

C a solution because salt is a solute

5 0

3 years ago

Read 2 more answers

Match each statement with one of these terms.

skelet666 [1.2K]

Answer:

1 - e, 2 - k, 3 - a, 4 - i, 5 - b,

Explanation:

The ratio of the amount of analyte in the stationary phase to the amount in the mobile phase. --- Retention factor.

Time it takes after sample injection into the column for the analyte peak to appear as it exits the column. -- Retention time

The process of extracting a component that is adsorbed to a given material by use of an appropriate solvent system. -- Elution

Measure of chromatographic column efficiency. The greater its value, the more efficient the column. -- Theoretical plate number

Gas, liquid, or supercritical fluid used to transport the sample in chromatographic separations. -- Mobile phase

Immiscible and immobile, it is packed within a column or coated on a solid surface. -- Stationary phase

8 0

2 years ago

1. Calculate the attractive bonding force for NaF and for MgO when the ions just touch. Based on the attractive bonding force, w

Andreas93 [3]

Answer:

The attractive force is negative and MgO has a higher melting point

Explanation:

From Couloumb's law:

Energy of interaction, E = k $\frac{q1q2}{r}$

where q1 and q2 are the charges of the ions, k is Coulomb's constant and r is the distance between both ions, i.e the atomic radii of the ions.

If you look at Coulomb's law, you note that in the force is negative (because q1 is negative while q2 is positive).

In addition to that, the compounds MgO and NaF have similar combined ionic radii, then we can determine the melting point trend from the amount of energy gotten

The melting point of ionic compounds is determined by 1. charge on the ions 2. size of ions. while NaF has smaller charges (+1 and -1), MgO (+2 and -2) has larger charges and greater combined atomic radii. This implies that the compound with greater force would have a higher melting point.

Hence the compound MgO would have a higher melting point than NaF.

5 0

3 years ago