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2 years ago

I need help fast answering these questions plzzz!! NO LINKS

Chemistry

1 answer:

Sindrei [870]2 years ago

4 0

Answer:

1. physical property

2. chemical property

3. physical change

hope this helps

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Please help! It would be amazing if you can!!

Anastaziya [24]

Answer:

3) Warm Temperatures and High pressures

Explanation:

This is the typical characteristic of southern air masses moving north

5 0

3 years ago

Determine the volume of occupied by 4 moles of carbon dioxide gas at

valentina_108 [34]

Answer:

Vol of 4 moles CO₂(g) at STP = 89.6 Liters

Explanation:

STP

P = 1 Atm

V =

T = 0°C = 273 K

n = 4 moles

R = 0.08206 L·Atm/mol·K

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (4 moles)(0.08206 L·Atm/mol·K)(273 K)/(1 Atm) = 89.6 Liters

5 0

3 years ago

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit

kati45 [8]

<u>Answer:</u> The $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

$CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}$

We are given:

$p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm$

Putting values in above equation, we get:

$K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = $5.34\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol$

Hence, the $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

5 0

3 years ago

8.5ml of a sample of sea water solution was added to a 44.317g evaporating dish the combination weighted 52.987g after evaporati

Ratling [72]

Answer: There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

5 0

2 years ago

5.80 gg of NaCl in 350. mL of a NaClsolution

Delicious77 [7]

Answer:

Solution is 0.28 M

You can also say, [NaCl] = 0.28 mol/L

Explanation:

As you have a solute mass and the solution's volume, you may find the molarity concentration of solution.

Molarity specifies the moles of solute in 1 L of solution

We convert the volume of solution to L → 350 mL . 1L / 1000 mL = 0.350L

We convert the mass of solute to moles → 5.80 g . 1mol / 58.45 g = 0.0992 moles

Molarity (mol/L) = 0.0992 mol /0.350L = 0.28M

6 0

2 years ago