All categories

3 years ago

Someone help answer this

Chemistry

2 answers:

Flura [38]3 years ago

7 0

Answer: Heat

Explanation:

Also considered as thermal energy. So when the person dives in water their heat temperature balances out.

alexandr402 [8]3 years ago

6 0

it is water energy :) this is easy

You might be interested in

The minerals hematite (Fe2O3) and magnetite (Fe3O4) exist in equilibrium with atmospheric oxygen:4Fe3O4(s) + O2(g) ⇌ 6Fe2O3(s)Kp

erastovalidia [21]

Answer:

For a: The partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

For b: The reaction must proceed in the forward direction to reach equilibrium.

For c: The value of $K_c$ is $6.12\times 10^{88}$

Explanation:

For the given chemical equation:

$4Fe_3O_4(s)+O_2(g)\rightleftharpoons 6Fe_2O_3(s)$

For a:

In the expression of $K_p$ , the partial pressures of solids and liquids are taken as 1.

The expression of $K_p$ for above equation follows:

$K_p=\frac{1}{p_{O_2}}$

We are given:

$K_p=2.5\times 10^{87}$

Putting values in above equation, we get:

$2.5\times 10^{87}=\frac{1}{p_{O_2}}\\\\p_{O_2}=\frac{1}{2.5\times 10^{87}}=4.0\times 10^{-88}atm$

Hence, the partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

For b:

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{1}{p_{O_2}}$

We are given:

$p_{O_2}=0.21$

Putting values in above equation, we get:

$Q_p=\frac{1}{0.21}=4.76$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction must proceed in the forward direction to reach equilibrium.

For c:

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 = $2.5\times 10^{87}$

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

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

T = temperature = 298 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=0-1=-1$

Putting values in above equation, we get:

$2.5\times 10^{87}=K_c\times (0.0821\times 298)^{-1}\\\\K_c=6.12\times 10^{88}$

Hence, the value of $K_c$ is $6.12\times 10^{88}$

7 0

2 years ago

Element X has two natural isotopes. The isotope with a mass of 10.012 amu has a relative abundance of 19.91%. The isotope with a

SIZIF [17.4K]

To find the atomic mass, multiply the individual atomic masses with their relative abundances and add them up.

10.012*0.1991 + 11.009*0.8009 = 10.810 amu (5 s.f.)

Element X is Boron.

4 0

3 years ago

A person observes that when the front porch of a house is wet , it is very slippery. The person adds sand to a paint can and rep

diamong [38]

Answer:

Explanation:

it will help stop frictoin

7 0

2 years ago

Identify each type of bond:

Sindrei [870]

Answer:

ionic bond=72%

covalent character=28%

Explanation:

3 0

3 years ago

Read 2 more answers

how much sodium chloride must be added to 100 mL of water so that its concentration is 20 parts per million assume that the dens

OlgaM077 [116]

Answer: 2mg

Part per million is unit that equal to mg/kg of water or simply $10^{-6}$ . In this case, you asked how much NaCl needed and know the volume of water(100mL), concentration(20ppm or 20 mg/kg) and water density.
Then, the equation would be

concentration = NaCl weight/ water weight
20 x $10^{-6}$ mg= NaCl weight/ (100ml x 1g/ml)
NaCl weight= 20x $10^{-6}$ mg/kg x 100g
NaCl weight= 2000 x $10^{-6}$ g
NaCl weight= 2x $10^{-3}$ g = 2mg

6 0

2 years ago