If you want to melt ice into liquid water, does the ice absorbs heat or release heat? Is this endothermic or exothermic?

Answers

Answer 1

Answer:

Answer:

Endothermic

Explanation:

This would be an endothermic reaction, due to the fact that the water would be signifigantly warmer then the ice cube, which causes it to melt.

Answer 2

Answer:

Answer:

absorbs heat, which makes it endothermic.

Explanation:

Basically melting ice is it in windows or Mac reaction makes the ice absorbs heat energy which causes a change to occur

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A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.pKb for the weak base = 4.82.

Molar mass of the weak base = 82.0343g/mole.

Note: pKa = -logKa

pKb = -logKb

pH + pOH = 14

[H+ ] [OH- ] = 10^-14

Answers

Answer:

11.39

Explanation:

Given that:

$pK_(b)=4.82$

$K_(b)=10^(-4.82)=1.5136* 10^(-5)$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (1.805\ g)/(82.0343\ g/mol)$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)$

$Molarity=(0.022)/(0.055)$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_(b)=\frac {\left [ BH^(+) \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136* 10^(-5)=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

The Aurora Borealis or northern light occur when solar flares react with the atmosphere near the north poles. The Aurora give off colred light because electricity is flowing. Which state of matter must the air be in for the Aurora to occur? A) gas B) liquid C) plasma D) solid

Answers

Answer: C) plasma

Natural phenomena characterized by colored lights in the sky, is caused by the interaction of charged particles of energy from the solar wind with the magnetic layer of the earth so we have turbulence and multicolored plasma clouds known as auroras arise.

Answer:

C) plasma

Explanation:

Plasma is the state where electrons are stripped allow these gases to conduct electricity.

.Which statement describes the difference between incomplete dominance and codominance?In codominance, only one allele is expressed in the offspring; In incomplete dominance, both alleles are expressed in the offspring.
In codominance, both allels are expressed in the offspring; In incomplete dominance, only one allele is expressed in the offspring.
In codominance, both alleles are expressed in the offspring; in incomplete dominance, the offspring demonstrate an intermediate form of the alleles from the parents.
In codominance, the offspring demonstrate an intermediate form of the alleles from the parents; in incomplete dominance, both alleles are expressed in the offspring.

Answers

Answer:

C Because im in 3rd grade and just did the test

Explanation:

im smart im in third grade but your doing this in college

Answer:

answer is c: In codominance, both alleles are expressed in the offspring; in incomplete dominance, the offspring demonstrate an intermediate form of the alleles from the parents.

Explanation:

Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solution
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution

Answers

Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.

To calculate the molarity (M) of a solution, you can use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

a. 0.38 moles of LiNO₃ in 6.14 L of solution:

Molarity (M) = 0.38 moles / 6.14 L = 0.062 M

b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:

First, you need to convert grams to moles using the molar mass of C₂H₆O.

Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol

Now, calculate moles of C₂H₆O:

moles = 72.8 g / 46.08 g/mol = 1.58 moles

Molarity (M) = 1.58 moles / 2.34 L = 0.675 M

c. 12.87 mg of KI in 112.4 mL of solution:

First, convert milligrams to grams (1 g = 1000 mg):

12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)

Now, convert mL to liters (1 L = 1000 mL):

112.4 mL = 0.1124 L

Calculate moles of KI:

Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol

moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles

Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M

So, the molarities of the solutions are as follows:

a. 0.062 M

b. 0.675 M

c. 0.000688 M

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Suppose of lead(II) acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Round your answer to significant digits.

Answers

Answer:

0.294 M

Explanation:

The computation of the final molarity of acetate anion is shown below:-

Lead acetate = Pb(OAc)2

Lead acetate involves two acetate ion.

14.3 gm lead acetate = Mass ÷ Molar mass

= 14.3 g ÷ 325.29 g/mol

= 0.044 mole

Volume of solution = 300 ml.

then

Molarity of lead is

= 0.044 × 1,000 ÷ 300

= 0.147 M

Therefore the molarity of acetate anion is

= 2 × 0.147

= 0.294 M

Final answer:

To calculate the final molarity of acetate anion in the solution, consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. When ammonium sulfate is added, it reacts with the lead(II) cations, leaving only the acetate anions in the solution. The final concentration of acetate anions is therefore the same as the initial concentration.

Explanation:

To calculate the final molarity of acetate anion in the solution, we need to consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. Lead(II) acetate will dissociate into lead(II) cations (Pb2+) and acetate anions (CH3COO-) in solution. However, when ammonium sulfate is added, the sulfate anions (SO42-) react with the lead(II) cations, forming lead(II) sulfate and removing them from solution. This leaves us with only the acetate anions.

First, calculate the concentration of the acetate anions in the lead(II) acetate solution. Then subtract the concentration of the acetate anions that reacted with the lead(II) cations to form lead(II) sulfate. This will give us the final concentration of acetate anions in the solution.

Let's assume we have an initial concentration of lead(II) acetate of X M. The dissociation of lead(II) acetate can be represented as:

Pb(CH3COO)2(s) ⇌ Pb2+(aq) + 2CH3COO-(aq)

Since we assume the volume of the solution doesn't change when the lead(II) acetate is dissolved, the initial concentration of acetate anions is also X M.

When ammonium sulfate is added, it reacts with the lead(II) cations according to the reaction:

Pb2+(aq) + SO4^2-(aq) ⇌ PbSO4(s)

Since the concentration of lead(II) sulfate is negligible, we can assume that all the lead(II) cations react with the sulfate anions. This removes the lead(II) cations from solution, leaving us with only the acetate anions.

Therefore, the final concentration of acetate anions is still X M.

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Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or negativea) Pb^2+(aq) + 2Cl-(aq) ---> PbCl2(s)b) CaCO3(s) ---> CaO(s) + CO2 (g)c) 2NH3(g) ---> N2(g) + 3H2(g)d) P4(g) + 5O2(g) ---> P4O10(s)e) C4H8(g) + 6O2(g) ---> 4CO2(g) + 4H2O(g)f) I2(s) ---> I2(g)

Answers

Answer: a) $Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$ : negative

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ : positive

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ : positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)$ : negative

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$ : positive.

f) $I_2(s)\rightarrow I_2(g)$ : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

As ions are moving to solid form , randomness decreases and thus sign of $\Delta S$ is negative.

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)$

As gas is changing to solid, randomness decreases and thus sign of $\Delta S$ is negative.

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

f) $I_2(s)\rightarrow I_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

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Which of the following statements best explains why the atomic radius increases from top to bottom of the periodic table? A. The electronegativity decreases from top to bottom of the periodic table, so the atoms near the bottom have an increased capacity for electrons. B. For the atoms lower in the periodic table, there is more difference in charge between the electrons and the protons, which allows electrons to orbit farther from nucleus. C. The ionization energy decreases from top to bottom of the periodic table, so the atoms near the bottom have an increased capacity for electrons. D. For the atoms lower in the periodic table, the valence electrons are in higher energy levels and farther from the nucleus.

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