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3 years ago

How would you use density to identify an unknown substance

Chemistry

2 answers:

alexdok [17]3 years ago

8 0

Seaport is correct for your question

ale4655 [162]3 years ago

4 0

You would measure the substances mass and its volume, then divide the mass by the volume.

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If the distance between a point charge and a neutral atom and is multiplied by a factor of 5, by what factor does the force on t

dexar [7]

Given :

The distance between a point charge and a neutral atom and is multiplied by a factor of 5.

To Find :

By what factor does the force on the neutral atom by the point charge change.

Solution :

We know, electrostatic force between two object is directly proportional to product of charge and inversely proportional to distance between them.

Now, charge in neutral atom is 0 C.

So, the electrostatic force between two of them is also 0 N.

Therefore, by changing distance between the charge the forces did no change ( it remains zero).

3 0

3 years ago

What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?

Neko [114]

To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol

When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.

5 0

3 years ago

When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I

Andreyy89

Answer:

S A T. NV

Explanation:

8 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Bi2(CO3)3<br> Name of this compound

Keith_Richards [23]

Answer:

The name of this compound is :

Bi2(CO3)3 = Bismuth Carbonate

Explanation:

The name of the compound is derived from the name of the elements present in it.

The rule followed while naming the compound are:

1. The first element (always the cation) is named as such .

2. The second element (The anion) end with "-ate , -ide ," etc

3. NO prefix is added while naming the first element.

For example : Bi2 can't be named as Dibismuth

Na2 = Can't be named as disodium

Hence the compound :

Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)

Bi = Bismuth

CO3 = carbonate

Bi2(CO3)3 = Bismuth Carbonate

The molecular mass of this compound is :

Molecular mass = 2 (mass of Bi) + 3(mass of C) + 6(mass of O)

= 2 (208.98)+3(12.01)+6(15.99)

= 597.987 u

5 0

3 years ago