Ask question

Ask question

All categories

4 years ago

7

5. How many moles are in 58.6 L of Nitrogen gas (N.) at STP?

1 answer:

pantera1 [17]4 years ago

6 0

Answer:

2.6166 moles

Explanation:

stp means: T=273k p=101.3kpa

R= 8.31(constant) v=58.6

using the ideal gas law; pv=nRt

101.3 x 58.6 = n x 8.31 x 273

n= (101.3 x 58.6)\(8.31 x 273)

n= 2.6166moles

You might be interested in

If the reactants on the left side of a chemical equation are C3H8 + 5O2, the products in a balanced equation could be

Basile [38]

Holy I'm not that smart lol but I'll get back to you

6 0

4 years ago

Read 2 more answers

Which branch science consist of?<br>

Rzqust [24]

Answer:

Modern science is typically divided into three major branches that consist of the natural sciences (biology, chemistry, physics, astronomy and Earth science), which study nature in the broadest sense; the social sciences (e.g. psychology, sociology, economics, history) which study people and societies; and the formal ...

7 0

3 years ago

Read 2 more answers

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can <u>put the values into the equation</u>:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can <u>solve for the initial temperature of gold</u>, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

Find the celcius equivalent of 133 degrees Kelvin

egoroff_w [7]

Answer:

<h2>406 K</h2>

Explanation:

To convert a temperature from degree Celsius to Kelvin add 273 to the value in degree Celsius

That's

K = 273 + °C

where

K is the temperature in Kelvin

°C is the temperature in degree Celsius

From the question we have

K = 273 + 133

We have the final answer as

<h3>406 K</h3>

Hope this helps you

5 0

3 years ago

Read 2 more answers

How does one recognize a redox reaction?

nika2105 [10]

Answer:

A.

Explanation:

A redox reaction is a reaction when oxidation states (or numbers) change during reaction.

3 0

3 years ago