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4 years ago

Can atoms compose molecules?

Chemistry

2 answers:

Nana76 [90]4 years ago

7 0

Answer: yes

Explanation: Atoms can join together to form a molecule

Naya [18.7K]4 years ago

5 0

Yes it forms in to molecules and if you want to know it can form into compounds

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Consider a solution that is 0.05 M HCl. Your goal is to neutralize 1 L of this solution (i.e. bring the pH to 7). You also have

Ilia_Sergeevich [38]

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of $HCl$ = Moles of NaOH

Considering :-

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

Given that:

$Molarity_{NaOH}=5\ M$

$Volume_{NaOH}=?\ L$

$Volume_{HCl}=1\ L$

$Molarity_{HCl}=0.05\ M$

So,

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

$0.05\times 1=5\times Volume_{NaOH}$

$Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L$

<u>The volume of NaOH required is - 0.01 L</u>

3 0

4 years ago

Which statements accurately describe binary star systems? Check all that apply. have more than two stars have stars that are alw

Westkost [7]

Answer:

have stars that might appear to wobble

often have one star that is brighter than the other

Explanation:

A binary star system is a star system made up of mostly two stars that moves round their common fixed center.

The two orbiting stars are gravitationally bonded to one another and they move round each other.

Most binary stars might appear wobble. One of the stars often appears brighter than the other.

6 0

3 years ago

Read 2 more answers

Platinum crystallizes in a face-centered cubic cell. the density of platinum is 21.4 g/cm3. calculate the radius of the platinum

IgorLugansk [536]

1 mole of platinum has a mass of 195 g therefore 1 atom will have a mass of 195 g /(6.02 ×10^23) = 3.239 × 10^-22 g
Density is given by dividing mass by volume, thus to get volume, mass is divided by density.
The volume = (3.239 × 10^-22)/21.4
= 1.514 × 10^-23 cm³
But volume of a sphere is given by 4/3πr³
Therefore, r³ = 3.6129 × 10^-24
r = ∛(3.6129 × 10^-24)
= 1.534 × 10^ -8 cm
Therefore, the radius of the platinum atom is 1.534 × 10^-8 cm

7 0

4 years ago

An analytical chemist has determined by measurements that there are 96.5 moles of carbon in a sample of acetic acid. how many mo

Nadya [2.5K]

The formula of acetic acid is CH3COOH => C2H4O2.

So, the acetic acid has the same number of atoms of carbon (C) than of oxygen (O).

Therefore, the sample that contains 96.5 moles of carbon, will contain also 96.5 moles of O.

Answer: 96.5 moles of oxygen.

3 0

3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

4 years ago