A student measured the masses of four different-sized blocks. The student determined that each block had a mass of 50 grams.


(There is a small block, a little bit bigger block, a big block and the biggles block)


Which block has the least density?

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Answer:

1.24 L of H₂ at STP .

Explanation:

2Al(s)    +6HCl(aq)    →    2AlCl₃(aq)   +    3H₂(g)

2 moles                                                  3 x 22.4 L

2 x 27 g of Al reacts to give  3 x 22.4 L of H₂ at STP .

1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .

= 1.24 L of H₂ at STP .

The volume of hydrogen produced by 1 grams of Al has been 1.24 L.

The balanced chemical reaction has been given as:

[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]

From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen

The mass of the compound from moles can be given as:

Mass = moles × molecular mass

Mass of 2 moles Al = 2 × 27 g

Mass of 2  moles Al = 54 g

Mass of 3 moles hydrogen = 3 × 2 g

Mass of 3 moles hydrogen = 6 g

From the equation,

54 g aluminum gives = 6 grams hydrogen

[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]

1 gram Aluminum = 0.11 grams hydrogen

The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:

Mass = moles × molecular mass

0.11 g = moles × 2 g/mol

Mole of hydrogen = 0.055 mol

The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.

According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,

1 mol = 22.4 L

0.055 mol = 0.055 × 22.4 L

0.055 mol = 1.244 L.

The volume of hydrogen produced by 1 grams of Al has been 1.24 L.

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Answer:

[tex]\boxed {\tt 1.0 \ g/mL}[/tex]

Explanation:

Density can be found by dividing the mass by the volume.

[tex]d=\frac{m}{v}[/tex]

The mass of the water is 50.0 grams.

The volume of the water is 50.0 milliliters.

[tex]m= 50.0\ g \\v=50.0 \ mL[/tex]

Substitute the values into the formula.

[tex]d=\frac{50.0 \ g}{50.0 \ mL}[/tex]

Divide.

[tex]d= 1.0 \ g/mL[/tex]

The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³

Answer:

Metals and Non-metals

Explanation:

Answer:

True

Explanation:

But have different neutrons

Answer:

C. three

Explanation:

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

[tex]2SO_2 + O_2 \to 2SO_3[/tex]

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction [tex]2SO_2 + O_2 \to 2SO_3[/tex] = 0.8325;

If we want to find:

[tex]SO_2 + \dfrac{1}{2}O_2 \to SO_3[/tex]

Then:

[tex]K_c = (0.8325)^{1/2}[/tex]

[tex]\mathbf{K_c = 0.912}[/tex]

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

Explanation:

its 399929939932929292

Answer:

0.2 moles of octane

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ------> 8CO2(g) + 9H2O(l)

From the equation;

1 mole of octane yields 9 moles of water

x moles of octane yields 1.4 moles of water

x= 1.4 × 1/9

x= 0.16 moles of octane

Answer:

they belong to same grop

Answer:

D) it is equal in magnitude but opposite in sign

Explanation:

The compounds (2S ,3S)-dibromopentane and (2R, 3R)-dibromopentane are a pair of enantiomers.

Enantiomers are stereoisomers that that are mirror images of each other and rotate plane polarized light. The magnitude of rotation of plane polarized light is known as specific rotation of the compound measured in degrees.

Enantiomers have the same magnitude of specific rotation but opposite in sign. This is  because one enantiomer rotates plane polarized light to the left while the other rotates plane polarized light to the right.

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

[tex]Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }[/tex]

Where [tex]Y_{O_{2} } \\[/tex] is the fractional oxygen saturation

[tex]{P_{O_{2} }[/tex] is the partial pressure of oxygen

[tex]P_{50}[/tex] is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

[tex]{P_{O_{2} }[/tex] = 40 mm Hg

[tex]P_{50}[/tex] = 22 mm Hg

h = 3

Putting these values into the equation, we get

[tex]Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }[/tex]

[tex]Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3} }[/tex]

[tex]Y_{O_{2} } = \frac{64000 }{10648 + 64000 }[/tex]

[tex]Y_{O_{2} } = \frac{64000 }{74648 }[/tex]

[tex]Y_{O_{2} } = 0.86[/tex]

Hence, the fractional saturation for hemoglobin is 0.86.

Answer:

D. Electrons move from Sphere A to Sphere B.

Explanation:

The statement that best describes the charging method is "Electrons move from Sphere A to Sphere B". This is further explained below.

A charge is simply defined as the property of a unit of matter that expresses how many more or fewer electrons it has than protons.

In conclusion, For a negatively charged rod placed near two neutral metal spheres, electrons move from Sphere A to Sphere B.

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Answer:

gold

Explanation:

Cations are positively charged, so the element that loses electrons to have the same amount of electrons as a noble gas atom is gold.

Answer:

See explanation

Explanation:

Hypatia is popular for her work in mathematics. She also did some work in the area of astronomy. Her well know work in mathematics is her ideas about conic sections.

She was born the Theon of Alexandria and she was a professional mathematician in her life time.

She was the greatest mathematician of her time and she was telling leader of the Neoplatonist school of philosophy in Alexandria. By so doing, she conquered the culture of sexism in her time.

She was trained by her father in mathematics and eventually replaced him. She was the last major mathematician in the Alexandrian tradition.

Answer:

false

Explanation:

Elements cannot be broken down into a simpler substance.

Answer:

None

Explanation:

We must recall that the number of hybrid orbitals formed must be equal to the number of atomic orbitals that were combined. Hence, if we have an sp3 hybridized orbital, there are four hybrid orbitals formed.

Since there are four hybrid orbitals formed from the three p and one s orbital that combined, there is no unhybridized orbital left.

The number of unhybridized 2p orbitals that remains on an atom that has sp3 hybrid orbital is : ( D ) None

The hybridization of an s orbital with three p orbitals will yield four sp3 hybrid orbitals, which means that there will be no unhybridized 2p orbitals left.

This is because during the hybridization, the number of hybrid orbitals yielded must be equal to the combined number of atomic orbitals that yielded the Hybrid orbitals.

Hence we can conclude that the number of unhybridized 2p orbitals that remain on an atom that has sp3 hybrid orbital is zero ( None )

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Answer:

telophase

Explanation:

Answer:

Explanation:

energy of solar radiation = 70 W / m²

energy absorbed in 1 hour by an area of .18 m²

= 70 x .5 x .18 x 60 x 60 J

= 22.68 x 10³ J

bond energy of i mole bond = 348 x 10³ J

bond energy of 6.02 x 10²³ bonds = 348 x 10³  J

bond energy of one bond = 57.8 x 10⁻²⁰ J  

No of bonds broken by energy 22.68 x 10³

= 22.68 x 10³  / 57.8 x 10⁻²⁰

= .3923 x 10²³

= 39.23 x 10²⁰ .

Answer:

There are approximately 11.0231 pounds in a kilogram

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

The standard enthalpy of the reaction is -827.5 kJ/mol

The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.

The given reaction can be expressed as:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The standard enthalpy can be represented by the equation:

[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]

At standard conditions, the standard enthalpies of formation of the given species are:

∴

[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]

[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]

[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]

Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol

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Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

Answer:

organ system

Explanation:

Answer:

B) organ

Explanation:

Answer:

2016 Km.

Explanation:

The following data were obtained from the question:

Speed (S) = 1.12×10⁴ m/s

Time (t) = 3 mins

Distance (d) =.?

Next, we shall convert 3 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

3 mins = 3 min × 60 s / 1 min

3 mins = 180 s

Next, we shall determine the distance travelled by the rocket. This is illustrated below:

Speed (S) = 1.12×10⁴ m/s

Time (t) = 180 s

Distance (d) =.?

Speed (S) = distance (d) /time (t)

S = d/t

1.12×10⁴ = d/ 180

Cross multiply

d = 1.12×10⁴ × 180

d = 2016000 m

Finally, we shall convert 2016000 m to Km. This can be obtained as shown below:

1000 m = 1 Km

Therefore,

2016000 m = 2016000 m × 1 Km / 1000 m

2016000 m = 2016 Km

Therefore, the rocket will travel 2016 Km in 3 mins

Answer:

D

Explanation:,

In designing an experiment there must be multiple trials. Multiple trials eliminate errors when the results are averaged.

This gives better results than when single measurements are used to determine results of experiments. The result of this experiment can be regarded as being reliable because it was well designed.

Answer:evaporation

Explanation:

Answer:

Pretty sure its evaporation, if its not I'm very sorry.

Answer:

2,2,3- trimethylhexane because it has more carbon atoms than pentane.

Explanation:

Hello.

In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.

In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.

Best regards.

Answer:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

Explanation:

We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.

Aqueous AgNO₃ and KI will dissociate in solution as follow:

AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)

KI (aq) —> K⁺(aq) + I¯(aq)

Aqueous AgNO₃ and KI will react as follow:

AgNO₃ (aq) + KI (aq) —>

Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)

Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.

Thus, the net ionic equation is:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

The net ionic equation of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex] to form [tex]AgI[/tex] precipitates is: B. [tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]

A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.

This ultimately implies that, both the charge on each atom and sum of the masses of the chemical compounds or elements in a chemical equation are properly balanced.

An ion can be defined as an atom or molecules (group of atoms) that has lost or gained one or more of its valence electrons, thereby, making it have a net positive or negative electrical charge.

First of all, we would write the dissociation equation for aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:

For [tex]AgNO_3[/tex]:

[tex]AgNO_3_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]

For [tex]KI[/tex]:

[tex]KI_{(aq)}[/tex] -----> [tex]K^{+}_{(aq)} + I^{-}_{(aq)}[/tex]

Next, we would write a chemical equation for the reaction of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:

[tex]AgNO_3_{(aq)} + KI_{(aq)}[/tex]  -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex] [tex]+ \;K^{+}_{(aq)} + I^{-}_{(aq)}[/tex] ----->[tex]AgI_{(s)} + K^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]

Note: Spectator ions refers to the ions that exist as a reactant and a product in a chemical equation because they are unchanged by the chemical reaction.

In this chemical reaction, the spectator ions are:

Finally, in order to obtain the net ionic equation, we would cancel out the two (2) spectator ions:

[tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]

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