A scientist is observing onion cells and human cheek cells under a microscope.

The Toucan has a long, narrow beak that allows it to reach fruit that is hard to reach for other birds. Plants, like the Monster

gladu [14]

<h2>Please mark me as brainliest please yaa!!!^_^</h2>

8 0

2 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

$deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol$

b) To calculate the constant we have the following expression:

$lnK=-\frac{deltaG_{rxn} }{RT}$

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

c) The equilibrium pressure of O₂ over M is:

$K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm$

3 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

Which Ion has the same electron configuration as an Ar atom

Troyanec [42]

Ar (argon) has 18 electrons

Cl- would give you 18 electrons
F- would give you 10 electrons
Li+ would give you 2 electrons
Na+ would give you 10 electrons

Cl- would be the correct answer

7 0

3 years ago

Which has more mass: an armful of feathers or an armful of lead? which has more volume: a kilogram of feathers or a kilogram of

valina [46]

Mass is related to weight. It measures the resistance of the substance to acceleration when a net force is applied.
Therefore: armful of lead has more mass

volume measures the amount of 3-dimensional space that an object occupies.
Therefore: kilogram of feather has more volume as it is less dense.

density is the mass per unit volume. Since armful of lead has more mass than a handful of feathers, while kilogram of feather has more volume than a kilogram of lead.
Therefore: kilogram of lead has higher density.

8 0

3 years ago