A skydiver slows down from 65 m/s to 5 m/s by opening the parachute. If this
takes 0.75 seconds, what is the skydiver's acceleration?
A. 45 m/s2 up
B. 80 m/s2 up
C. 45 m/s2 down
D. 80 m/s2 down

Answer:

Half life = 13.197 hour

Explanation:

Given:

Old amount (A₀) = 3.2

New amount (A) = 0.4

Radiation decay time (t) = 39.6 hour

Half life = T(1/2)

Find:

Half life = T(1/2) = T

Computation:

A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]

[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2

-[27.4428 / T] = In (0.125)

-[27.4428 / T] = -2.0794

[27.4428 / T] = 2.0794

T = 13.197

Half life = 13.197 hour

Answer:

ΔT=[tex]0.87^{\circ}C[/tex]

Explanation:

Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):

ΔT=[tex]Kb*m[/tex]

Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).

Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:

[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]

Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:

[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]

Finalmente para el calculo de la molalidad podemos dividir los dos valores:

[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]

Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:

ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]

Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:

Temperatura final = 100 + 0.87 = 100.87 ºC

Espero sea de ayuda!

Answer:

35 hrs

Explanation:

half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr

initial amount [tex]N_{0}[/tex] = 75 g

The final amount [tex]N[/tex] = 3.9 g

The time elapsed [tex]t[/tex] = ?

we use the relationship

[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]

substituting values, we have

3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

take the log of both side

log 0.052 = log  [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2

-1.284 =  [tex]\frac{t}{8.1}[/tex]  x -0.301

1.284 = 0.301t/8.1 =

1.284 = 0.0372t

t = 1.284/0.037 = 34.5 ≅ 35 hrs

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

Answer:

37.8

Explanation:

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

Moles of KHP =

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Volume of the calcium hydroxide solution used is 0.0235 mL.

Answer:

.0027 M

Explanation:

We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:

Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)

Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2

[PO3−4]=0.0027 M

Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.

Answer:

a

Explanation:

answer is a on edg

Answer:

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Explanation:

Consider the reaction:

H2CO(g) + O2(g) --------> CO2(g) + H2O(l)

Using standard thermodynamic data;

Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol

At standard thermodynamic data

[tex]\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}[/tex]

[tex]S^0(CO_2)[/tex] = 213.79 J/mol.K

[tex]S^0(H_2O)=[/tex]  69.95  J/mol.K

[tex]S^0 ({H_2CO}) =[/tex] 218.95 J/mol.K

[tex]S^0 (O_2)[/tex]  = 205.2 J/mol.K

[tex]\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K - (218.95+ 205.2) J/mol.K}[/tex]

[tex]\mathtt{S^0_{rxn} = (283.74) J/mol.K - (424.15) J/mol.K}[/tex]

[tex]\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}[/tex]

Based upon the stoichiometry of the reaction the sign of Sºrxn should be negative

Explanation:

Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

Answer:

A) 3.17 g of Zn

Explanation:

Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.

Zn⁺²(aq) + 2e⁻ → Zn(s)

We can establish the following relations:

The mass of Zn deposited under these conditions is:

[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]

Answer:

A.) 3.17

Explanation:

I got it right in class!

Hope this Helps!! :))

Answer:

Caesium reacts rapidly with water to form a colourless basic solution of caesium hydroxide (CsOH) and hydrogen gas (H2).

Explanation:

The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.

Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

Learn more about moles here: https://brainly.com/question/24817060

Answer:

176.8 g of ammonia, NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6 g

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34 g.

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.

This can be obtained as follow:

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.

Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.

Answer:

Molecular mass of HNO₃ = 63.015 g/mol

Explanation:

Note: The given question is incomplete , the given substance is nitric acid

Given:

Nitric acid (HNO₃)

Find:

Molecular mass

Computation:

Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)

Molecular mass of HNO₃ = 1.008 + 14.007 + 48

Molecular mass of HNO₃ = 63.015 g/mol

Answer:

Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.

Explanation:

Answer:

[tex]\Delta G=-359\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:

[tex]\Delta G=\Delta H -T\Delta S\\[/tex]

In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:

[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]

Best regards.

Answer:

B- 358 kj

Explanation: I took the test

Answer:

i believe its 42

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

You can learn more about neutralization here: https://brainly.com/question/16255996

Answer:

a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

b. 18.17% of Fe in the sample

Explanation:

a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:

Fe²⁺ → Fe³⁺ + 1e⁻

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

5 times the iron and suming the manganese reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

b. Moles of permanganate in the titration are:

0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻

Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:

1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =

4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:

4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =

Percent mass is:

0.2474g Fe / 1.362g sample ₓ 100 =

The mass percent of iron in the sample is 22.6%.

The net ionic equation of the reaction is;

5Fe^2+(aq) + 8H^+(aq) + MnO4^-   -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)

Number of moles of MnO4^-  = 39.42/1000 L × 0.0281 M = 0.0011 moles

If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-

x moles of Fe^2+ reacts with 0.0011 moles

x =  5 moles × 0.0011 moles/1 mole

x = 0.0055 moles

Mass of Fe^2+ =  0.0055 moles × 56 g/mol = 0.308 g

Mass percent of iron = 0.308 g/ 1.362 g × 100/1

= 22.6%

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The name of a compound is determined by IUPAC nomenclature

The International Union of Pure and Applied Chemistry have designated a universal rule for the nomenclature of organic compounds generally known as IUPAC nomenclature.

The whole idea of the IUPAC nomenclature is to have a universally agreed pattern of naming organic compounds according to their structure.

The compound is always named in such a way that the substituents and the functional group receives the lowest number.

In this case, the functional group is alkyne hence the name of the compound ends in -yne. The methyl group is substituted on position four.

Putting all these together the compound should be named as;  4-methylhept-2-yne

For IUPAC nomenclature of an inorganic compound

see:

https://brainly.com/question/11587934

Answer:

70 W and 40 w

there are 28 days in a month

Add those 2 together and get 110

then mutlpy by 28

3,080 amount of energy

Answer:

All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

A 1.89 pints of blood would contain 873 grams of platelets.

To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:

                                  [tex]1.89 \times 473.2 = 894.3 mL[/tex]

                                         [tex]\frac{894.3}{1000}= 0.84L[/tex]

Now, just calculate the amount of platelets present in 0.84L:

                                    [tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]

                                       x = 873 grams

So, a 1.89 pints of blood would contain 873 grams of platelets.

Learn more about transformation of units in: brainly.com/question/10667910

Answer:

C) formaldehyde, H2C=O.

Explanation:

Hello,

In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).

Best regards.

Answer:

False because it is false