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2 years ago

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Water contracts as it freezes at 0°C. Solid ice is less dense than liquid water. Liquid water expands with increasing temperatur

e between 0°C and 4°C. Liquid water expands with increasing temperature above 4°C.

1 answer:

Mariulka [41]2 years ago

6 0

Answer:

Explanation:no when water frezzing it will expand and when it weams up it shunkes

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Which statement best describes the effect of low ionization energies and low electronegativities on metallic bonding?

kramer

The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
Due to these low energies and negativities, valence electrons can be moved around quite easily and their positions may be altered quite drastically.

6 0

2 years ago

Read 2 more answers

How to prepare 250.00 ml of approximately 1.0 m hcl solution from the 2.5 m hcl solutio?

kirill115 [55]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2.5 M x V1 = 1.0 M x .250 L

<span>V1 = 0.10 L or 100 mL of the 2.5 M HCl solution is needed

Hope this helps.</span>

4 0

3 years ago

Read 2 more answers

To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water

Masteriza [31]

Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0

3 years ago

The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em> A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em> </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

<em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

Feasible. ΔG°rxn = ΔG°f(product) < 0.

<em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

<em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0

3 years ago

the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of B

ddd [48]

Answer:

1.8 × 10⁻⁴ mol M/s

Explanation:

Step 1: Write the balanced reaction

2 Br⁻ ⇒ Br₂

Step 2: Establish the appropriate molar ratio

The molar ratio of Br⁻ to Br₂ is 2:1.

Step 3: Calculate the rate of appearance of Br₂

The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:

3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s

3 0

2 years ago