Ask question

Ask question

All categories

Harlamova29_29 [7]

2 years ago

5

Why do heated molecules rise? a Because they are less dense than the molecules around them b Because they are moving slower than

the other molecules c Because they are denser than the molecules around them

1 answer:

Volgvan2 years ago

7 0

Answer:

A because they are less dense than the molecules around them

You might be interested in

A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of

castortr0y [4]

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains $1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$ .

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$ .

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.

6 0

3 years ago

18 ) Liquid A and Liquid B are clear liquids. They are placed in open containers and allowed to evaporate. When evaporation is c

arsen [322]

Answer:

Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture

Explanation:

5 0

3 years ago

There are 55.0 g of neon gas in a 3.0 L cylinder, the pressure is 4.5 atm what is the temp of the gas

marshall27 [118]

Answer:

60.9 Kelvin

Explanation: First, write out everything that you know. You are tring to find the temperature, so the temperature will be represented by x.

Pressure (P)= 4.5 atm

Volume (V)= 3L

Number of Moles (n)= ?

Gas Consant (R)= 0.0821, if the pressure is in atm, that means r is 0.0821

Temperature (T)= x

We don't have all the information we need to plug the values into the equation. We still need to know how many moles 55.0 grans of neon is.

Ne in Grams= 55

Atomic Mass of Ne= 20.1797

55/20.1797= 2.7

moles= 2.7

Now that we have all the information we need, plug everying into the equation. In case you don't know, the Ideal Gas Law Equation is PV= nRT.

(4.5)(3) = (2.7)(0.821)x

x= 60.9

Now you have your temperature! It is 60.9 in Kelvin.

4 0

2 years ago

How many grams of butane were in 1. 000 atm of gas at room temperature?

dimaraw [331]

The mass in grams of butane at standard room temperature is 53.21 grams.

<h3>How can we determine the mass of an organic substance at room temperature?</h3>

The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:

PV = nRT

Pressure = 1.00 atm
Volume = 22.4 L
Rate = 0.0821 atm*L/mol*K
Temperature = 25° C = 298 k

1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)

n = 22.4/24.4658 moles

n = 0.91556 moles

Recall that:

number of moles = mass(in grams)/molar mass

mass of butane = 0.91556 moles × 58.12 g/mole

mass of butane = 53.21 grams

Learn more about calculating the mass of an organic substance here:

brainly.com/question/14686462

#SPJ12

3 0

2 years ago

How many molecules are in 100 g of C6H120,?*

Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0

3 years ago