All categories

2 years ago

1.52 moles of gas are at a pressure of 99.5kPa and a temperature of 298K. What is the volume of the gas

Chemistry

1 answer:

dangina [55]2 years ago

7 0

Answer:

Explanation:

answer my last question plz

You might be interested in

A student is learning about a particular chemical element. The element is a metal that readily forms ionic salts with nonmetals

seropon [69]

Answer:

Group 2

Explanation:

Because group 2 is very reactive and forms salts with other non metals. Example Calcium Chloride. Calcium is from group 2, forms ionic salts with other non metal, and exists as solid at standard temp and pressure.

3 0

3 years ago

Read 2 more answers

Why do you think DNA has had such an impact on forensic science

Nuetrik [128]

Every single person has their own unique DNA so with this, if they find any fingerprints they'll be able to track them down easily

8 0

3 years ago

Read 2 more answers

What is a change that will not affect the pressure in a container?

allochka39001 [22]

Answer:

Density of the contents

Explanation:

7 0

3 years ago

Convert the following temperatures to Kelvin:

maxonik [38]

Explanation:

A. 100°C to Kelvins

$T(K)=T(^oC)+273.15$

$T(K)=100(^oC)+273.15=373.15 K$

B 600°R to Kelvins

$(T)^oK=((T)^oR)\times 1.8$

$(T)^oK=600\times 1.8 K = 1080 K$

C. 98°F to Kelvins

$(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}$

$(T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K$

D. 77.4°F to degree Celsius

$((T)^oC)=((T)^oF-32)\times \frac{5}{9}$

$(T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC$

E. 77.4 K to degree Celsius

$T(^oC)=T(^K)-273.15$

$T(^oC)=77.4(K)-273.15=-195.75^oC$

F. 77.4°R to degree Celsius

$(T)^oC=((T)^oR-491.67)\times \frac{5}{9}$

$(T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC$

7 0

3 years ago

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

Veseljchak [2.6K]

Answer:

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

$CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol$

Mass of methane burnt = 530 g

Moles of methane burnt = $\frac{530 g}{16 g/mol}=33.125 mol$

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

$-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ$

$-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ$

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

7 0

3 years ago