3. The dry air in the Alps helped to prevent damage to the cells in the Iceman's body.

List 5 of the signs that can identify the evidence of a chemical change.

trapecia [35]

Answer:

here are some ideas <3

Explanation:

burning of paper.

cooking of food

burning of wood

rotting of fruits.

frying egg

rusting of iron

mixing acid and base.

burning of candle

leaves changing color

melting of sugar.

baking

explosion of fireworks.

souring milk

digestion of food

fermentation

lighting matchstick

photosynthesis

decomposition of waste

8 0

3 years ago

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric che

Helga [31]

Answer:

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

Explanation:

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

Initially

3.0 atm 0

At equilibrium

(3.0-2p) p

Equilibrium partial pressure of $NO_2=2.1atm=3.0-2p$

p = 0.45 atm

The value of equilibrium constant wil be given by :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}$

$K_p=\frac{0.45}{(2.1)^2}=0.10$

After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

After adding 1.5 atm of $NO_2$ :

(2.1+1.5) atm 0.45 atm

At second equilibrium:'

(3.6-2P) (0.45+P)

The expression of equilibrium can be written as:

$K_p=\frac{p'_{N_2O_4}}{(p'_{NO_2})^2}$

$0.10=\frac{(0.45+P)}{(3.6-2P)^2}$

Solving for P:

P = 0.37 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time:

= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

5 0

3 years ago

Write the balanced nuclear equation for alpha decay of polonium−218. include both the mass numbers and the atomic numbers with e

Dvinal [7]

Alpha particle is nucleus of a helium-4 atom, which is made of two protons and two neutrons.
Nuclear reaction: ₈₄²¹⁸Po → ₈₂²¹⁴Pb + α (alpha particle).
Alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.

7 0

3 years ago

How can you identify balanced forces

solmaris [256]

Answer:

To determine if the forces acting upon an object are balanced or unbalanced, an analysis must first be conducted to determine what forces are acting upon the object and in what direction. If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced.

3 0

3 years ago

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

3 years ago