All categories

3 years ago

Classify these substances as elements, mixtures or compounds. gold dust - honey - ink

2 answers:

8 0

<span>Honey - mixture </span>
<span>Ink - mixture </span>
<span>Gold dust - compound </span>

Hope that helps! =3
Have a Blessed rest of your night! =3

Musya8 [376]3 years ago

6 0

Honey is a - Mixture

Ink is - Mixture

And Gold Dust is a - Compound

You might be interested in

Determine the number of moles of H in each sample

MissTica

C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole

D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles

<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>

1 mole of N₂H₄ contains 4 moles of H

Therefore,

0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H

<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>

1 mole of C₁₀H₂₂ contains 22 moles of H

Therefore,

34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H

Learn more about mole:

brainly.com/question/13314627

#SPJ1

5 0

2 years ago

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

1. Describe how SHAPE can change during a collision. Be specific with your evidence and add detail to your answer.

a_sh-v [17]

Answer:

Hey

of course, the damage of a collision depends upon how fast to objects are moving at each other and how strong they are. If you have two tanks moving at each other 2 miles per hour it will be very little damage and the ->shape<- will not change much, maybe a dint or two. But if two balloons filled with water are moving at each other 5000 mph they will completely evoporate in a burst of light, and their ->shape<- will change very much. This is how shape and motion are related.

Hope it helped

spiky bob your answerer

4 0

3 years ago

.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f

Tasya [4]

Answer : The value of $K_c$ for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)$

The expression of $K_c$ will be,

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

First we have to calculate the concentration of $Br_2,Cl_2\text{ and }BrCl$ .

$\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M$

Now we have to calculate the value of $K_c$ for the given reaction.

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

$K_c=\frac{(0.6)^2}{(1)\times (1)}$

$K_c=0.36$

Therefore, the value of $K_c$ for the given reaction is, 0.36

6 0

3 years ago

Read 2 more answers

Name the type of organic compound from each description of the functional group: (b) group that is not polar and has a double bo

krek1111 [17]

Some of the crucial functional groups in organic compounds are the hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl groups. The formation of molecules including DNA, proteins, carbohydrates, and lipids depends heavily on these groups.

<h3>What is Function Group ?</h3>

In organic chemistry, a functional group is a collection of atoms within molecules that interact to cause predictable reactions. Functional groups include the hydroxyl, ketone, amine, and ether groups, for instance.

(b) Most double bonds, such those found in alkenes, are formed when two carbon atoms are present. In a carbonyl group, for instance, where a carbon atom and an oxygen atom are together, there are several double bonds connecting the two separate components. Sulfoxides (S=O), imines (C=N), and azo compounds (N=N) also include common double bonds.

To know more about Functional Group please click here : brainly.com/question/493841

#SPJ4

8 0

2 years ago