I took 100 micrograms of vitamin d yesterday (13yrs old) is it bad

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

3 years ago

Explain how a drought poses a threat to life and the biosphere. Use complete sentences.

Marina CMI [18]

Answer:

A drought poses a huge threat to all life. If a drought occurred the entire food chain would disintegrate within months. There would be no water for any animals or plants. Small mammals would not be able to eat plants, and reptiles would not be able to the small animals, and so on. A drought can destroy an ecosystem in a short amount of time.

Water is only being moved into the air through water vapor, so the air will become hotter than water. Land is also not moving in the atmosphere and absorbing heat like air is, so air will also be hotter than land, depending on what the land is made of.

Through precipitation, water in the atmosphere can return to the hydrosphere or percolate into the ground to become groundwater—part of the geosphere. ... Water in the biosphere can be released into the atmosphere through transpiration in plants, or respiration in animals.

Explanation:

6 0

2 years ago

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Drops of a clear liquid are added to a red liquid. As the drops are added, the red liquid becomes orange, then yellow

lakkis [162]

This must be a universal indicator, the pH is going down everytime you add more :)

3 0

3 years ago

Read 2 more answers

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago