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luda_lava [24]
3 years ago
7

Which has a higher electronegativity phosphorus or sodium

Chemistry
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

it is phosphorus it has o most 2 times as much  electromagnetically

Explanation:

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Does anyone have the answer sheet for all of Cell energy cycle gizmo part c?
FinnZ [79.3K]

Answer:

No but i wish

Explanation:

7 0
3 years ago
Exactly one mole of an ideal gas is contained in a 2.00-liter container at 1,000 K. What is the pressure exerted by this gas?
levacccp [35]
Use PV =nRT

so P = nRT/V

= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L

= 41 atm
7 0
3 years ago
Read 2 more answers
Which of the following statements does not describe what he structure of an atom an atom has a small dense center called nucleus
nikklg [1K]

Answer:

inside the nucleus of an atom are protons and electrons.

5 0
3 years ago
An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is
Juli2301 [7.4K]

Answer:

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}  

Explanation:

The gas ideal law is  

PV= nRT (equation 1)

Where:

P = pressure  

R = gas constant  

T = temperature  

n= moles of substance  

V = volume  

Working with equation 1 we can get  

n =\frac{PV}{RT}

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

\frac{m}{mw} =\frac{PV}{RT}  or  

m =\frac{P*V*mw}{R*T}   (equation 2)

The cylindrical container has a constant pressure p  

The volume is the volume of a cylinder this is

V =(pi)*r^{2}*h

Where:

r = radius  

h = height  

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is  V =(pi)*r^{2}*h

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:  

T =\frac{T_{1} + T_{O}}{2}  

Replacing these values in the equation 2 we get:

m =\frac{P*V*mw}{R*T}   (equation 2)

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}    

8 0
3 years ago
Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0
3 years ago
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