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GarryVolchara [31]
4 years ago
14

Which two chemists organized elements based on properties such as how the elements react or whether they are solid or liquid?

Chemistry
2 answers:
Alenkinab [10]4 years ago
8 0

Lavoisier and Döbereiner

I took the test

nataly862011 [7]4 years ago
7 0

Answer:

Dmitri Mendeleev and Henry Moseley developed their periodic tables about forty years apart. During that time, many discoveries in chemistry were made.

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Explain why the weight of an object is different on Earth and the moon even though the object’s mass is the same in both places.
Vlad [161]
There is a thing called gravity. Gravity in the earth causes anything to stay down. There was an experiment a person held a plastic box with bricks on earth. It was really hard to pick it up. Then in the ISS (International Space Station) the same person help and it was so much easier no hardship at all, even a baby can do it!! 
5. Length times With times Hight = Volume or L*W*H=V
4 0
3 years ago
Read 2 more answers
Particles in a are fixed in place and cannot move. They around a fixed position.
jonny [76]

Answer:

Solids have a set shape and can't flow since their particles can only vibrate in a specific direction. They are unable to travel from one location to another.

Explanation:

5 0
3 years ago
. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w
Ivahew [28]

Answer:  

(a) 45.17×10^-14 kg/s  

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

Explanation:  

Helium gas at temperature T=293k  

Helium gas at pressure P= 500kPa  

The inner diameter of spherical tank is D_1 = 2m  

The inner radius of spherical tank is : r_1 = \frac{D_1}{2}  

= \frac{2}{2}  

=1m  

Thickness of the container r = 1cm =0.01m  

Outer radius of the spherical tank is ;  

t = r_2 - r_1  

-r_2 = -t - r-1  

multiplying through with (-) we have ;  

r_2 = t + r_1  

r_2 = 1 + 0.01m  

r_2 = 1.01m  

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as  

D_A_B =4.0 ×10^-14 \frac{m^2}{s}  

From table molar mass and gas constant, the molecular weight of helium is:

 

M = 4.003kg/kmol  

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.  

S_He = 0.00045 kmol/m³. bar  

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as  

C_B_I = S_H_e×P  

= 0.00045kmol/m³. bar × (5)  

C_B_I = 2.25×10^-3 kmol/m³  

From one dimensional mass transfer through spherical layers is expressed as:

N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}

substituting all the values in the above relation, we have;

M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}

M_di_f_f=11.42×10^-14kmol/s

(a) The mass flow rate is expressed as

M_di_f_f = MN_diff

M_di_f_f=4.003×11.42×10^-14

M_di_f_f=45.71×10^-14kg/s

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

N_di_ff =11.42×10^-14kmol/s

N_di_ff= 11.42×10^-14×7×24×3600 kmol/week

N_di_f_f=6.9×10^-8kmol/week

The volume of the spherical tank is V=\frac{4}{3} πr_1^3

V=\frac{4}{3}π×1^3

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week

N_di_f_f-final ≅ 0.86

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

3 0
3 years ago
How many grams of acetylene are produced by adding 3 moles of cac2
jolli1 [7]

Answer:

78.12g of acetylene

Explanation:

Acetylene and calcium hydroxide are produced when H₂O reacts with CaC₂. The reaction is:

2 H₂O + CaC₂ → C₂H₂ + Ca(OH)₂

<em>Where 1 mole of C₂H₂ (acetylene) is produced per mole of CaC₂</em>

Thus, the addition of 3 moles of CaC₂ produces 3 moles of acetylene.

As molecular mass of acetylene is 26.04g/mol, grams of acetylene produced are:

3mol C₂H₂ × (26.04g / mol) = <em>78.12g of acetylene</em>

8 0
3 years ago
The equation for another reaction used in industry isCO(g) + H₂O(g) <img src="https://tex.z-dn.net/?f=%5Crightleftharpoons" id="
Sloan [31]

Answer:

(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4

(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol

Explanation:

(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol    3.2 mol      0          0              <em>Initial</em>

-x              -x                +x        +x            <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x       3.2-x            x           x             <em>Equilibrium</em>

In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol

The amounts of CO and H₂O are:

CO = 2.0 - 1.6 = 0.4 mol

H₂O = 3.2 - 1.6 = 1.6 mol

The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:

Kc = (1.6x1.6)/(0.4x1.6)

Kc = 1.6/0.4

Kc = 4

(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol  2.0 mol      0          0                 <em>Initial</em>

-x              -x             +x         +x               <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x        2.0-x         x           x                <em>Equilibrium</em>

Kc = (x*x)/((2.0-x)*(2.0-x))

4 = x²/(4 - 4x + x²)

16 - 16x + 4x² = x²

3x² - 16x + 16 = 0

Using Baskhara's equation:

Δ =(-16)² - 4x3x16

Δ = 256 - 192

Δ = 64

x = (-(-16) +/- √64)/(2*3)

x' = (16 + 8)/6 = 4

x'' = (16 - 8)/6 = 1.33

x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.

5 0
3 years ago
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