Ask question

Ask question

All categories

3 years ago

9

A gas with a volume of 300 mL at 150 °C is heated until its volume is 600 mL. What is the new temperature of the gas if the pres

sure is unaltered?

1 answer:

Mandarinka [93]3 years ago

3 0

.012K

Use the equation of $\frac{V1}{T1}$ = $\frac{V2}{T2}$ .

Replace V with volume and T with the temperature. However, with the temperature, make sure to convert the celsius into kelvin by adding 273.

Hope this helps!

You might be interested in

when 10 moles of sodium reacts with excess oxygen for 60 minutes, only 3 moles of sodium remains in 1000 millimeters of H20. Wha

diamong [38]

That would be....................

8 0

2 years ago

How are biology and Chemistry are related give any two points or reasons

k0ka [10]

Answer:

Both of them are related most of the structures of human body .The working of cells, the cellular respiration processes ,the proteins synthesizing carbohydrates syntheses lipids synthesis and many other reactions involves chemistry.

•Chemistry and biology both are branches of science so they are much more related to each other.

Explanation:

<h2>I hope it's helpful for you</h2>

6 0

2 years ago

Read 2 more answers

Which isotope will spontaneously decay and emit particles with a charge of +2?

VikaD [51]

Answer: The correct option is 4.

Explanation: All the options will undergo some type of radioactive decay processes. There are 3 decay processes:

1) Alpha decay: It is a decay process in which alpha particle is released which has has a mass number of 4 and a charge of +2.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

2) Beta-minus decay: It is a decay in which a beta particle is released. The beta particle released has a mass number of 0 and a charge of (-1).

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

3) Beta-plus decay: It is a decay process in which a positron is released. The positron released has a mass number of 0 and has a charge of +1.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta$

For the given options:

Option 1: This nuclei will undergo beta-plus decay process to form $_{25}^{53}\textrm{Mn}$

$_{26}^{53}\textrm{Fe}\rightarrow _{25}^{53}\textrm{Mn}+_{+1}^0\beta$

Option 2: This nuclei will undergo beta-minus decay process to form $_{80}^{198}\textrm{Hg}$

$_{79}^{198}\textrm{Au}\rightarrow _{80}^{198}\textrm{Hg}+_{-1}^0\beta$

Option 3: This nuclei will undergo a beta minus decay process to form $_{56}^{137}\textrm{Ba}$

$_{55}^{137}\textrm{Cs}\rightarrow _{56}^{137}\textrm{Ba}+_{-1}^0\beta$

Option 4: This nuclei will undergo an alpha decay process to form $_{85}^{216}\textrm{At}$

$_{87}^{220}\textrm{Fr}\rightarrow _{85}^{216}\textrm{At}+_2^4\alpha$

Hence, the correct option is 4.

7 0

3 years ago

Read 2 more answers

Select the reagents you would use to synthesize the compounds below from benzene. Use the minimum number of steps.

iris [78.8K]

Answer:

A

C

Explanation:

For the reaction of m-bromonitrobenzene from benzene first, we need to carry out the nitration of benzene utilizing HNO3, H2SO4.

Bromination of nitrobenzene with Br2, FeBr3 outfits m-bromonitrobenzene in light of the fact that nitro group is meta coordinating for electrophile. The mechanism activity for the reaction is shown in the image below.

5 0

2 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

irinina [24]

Answer:

$m_{KBr}=6.030gKBr$

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

$(T_b-T_b_0)=i*m*K_b$

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

$m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\ \\m=0.507mol/kg$

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

$n_{KBr}=0.507mol/kg*0.1kg=0.0507mol$

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

$m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr$

Best regards.

3 0

3 years ago