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2 years ago

What is the 2nd step in the process of natural selection?

Chemistry

1 answer:

Nesterboy [21]2 years ago

3 0

Answer: Selection proper

Explanation:

it's an anti-chance process, but subject to many constraints

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Calculate the mass in grams of 2.56 × 10 −3 mol of iron.

gizmo_the_mogwai [7]

Answer:

1.43 grams

Explanation:

Fe = 55.8 grams Fe = 1 mole Fe

2.56 • 10^-3 moles Fe / 1 • 55.8 grams Fe / 1 mole Fe = 1.43 grams Fe

Basically, you're just multiplying the molar mass of Fe (iron) by the moles of 2.56 • 10^-3 Fe, to find how many grams are in it.

2.56 • 10^-3 moles Fe = 1.43 grams Fe

5 0

3 years ago

How much heat is required to melt 26.0 g of ice at its melting point?

MAXImum [283]

Answer:

Heat required to melt 26.0 g of ice at its melting point is 8.66 kJ.

Explanation:

Number of moles of water in 26 g of water: 26× $\frac{1}{18.02}$ moles

=1.44 moles

The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol.

we have relation as:

q = n × ΔH

where:

q = heat

n = moles

Δ H = enthalpy

So calculating we get,

q= 1.44*6.02 kJ

q= 8.66 kJ

We require 8.66 kJ of energy to melt 26g of ice.

8 0

3 years ago

What is the mass because I already have the volume and density

Bad White [126]

76.88 I think im sorry if wrong

5 0

3 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

Be sure to answer all parts. For each of the following pairs of elements, state whether the binary compound they form is likely

andrew-mc [135]

Answer:

(a) Covalent bond. NF₃ (nitrogen trifluoride)

(b) Ionic bond. LiCl (lithium chloride)

Explanation:

Nitrogen and fluorine are nonmetals, with high and similar electronegativities, so they form covalent bonds, in which they share pairs of electrons to complete the octet in their valence shell. N has 5 valence electrons so it will form 3 covalent bonds while each Cl has 7 valence electrons so it will form 1 covalent bond. As a result, the empirical formula is NF₃ (nitrogen trifluoride).

Lithium is a metal and Chlorine is a nonmetal. They have different electronegativities so they form an ionic bond, in which Cl gains 1 electron (7 valence e⁻) and Li loses 1 electron (1 valence e⁻). The empirical formula is LiCl (lithium chloride).

5 0

3 years ago