5. Express the number 1.36 x 104 in common decimal form.

What is the composition of 'Clean dry air' - Idk what subject its just science

Luba_88 [7]

Dry air is a mixture of nitrogen, oxygen, carbon dioxide etc.
air is a mixture of gases 78% nitrogen an 21% oxygen and other components.

7 0

3 years ago

Asap! first to answer correctly gets 10 pts plus brainliest! :)

olga_2 [115]

Answer:

c

Explanation:

c first quarter waxing half moon

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

How many grams of h3po4 are in 255 ml of a 4.50 m solution of h3po4?

murzikaleks [220]

H3PO4 has molecular weight of approximately 98 grams per mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times 4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I automatically equate 1 Liter to 1000 mL since the given volume is in mL for easier computation.

7 0

3 years ago

The radioactivity of U-235 is of low intensity. Why then were the people of Hiroshima exposed to high intensity radiation in the

Marina86 [1]

Answer:

Explanation:

What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.

7 0

3 years ago