Determine which base will work to deprotonate each compound in an acid/base extraction.

Benzene ring with a carboxylic acid on one carbon and a methyl ester group on the opposite carbon __________
Benzene ring with a three carbon chain attached to one carbon. At the other end of the three carbon chain is a hydroxyl group. ___________
Benzene ring with a hydroxyl group attached to one carbon and a tertbutyl group attached to the opposite carbon __________-

A 500.0 g sample of aluminum, initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. 32.62245 kJ heat has been absorbed by the sample.

The term specific heat is defined as the amount of heat required to increase the temperature of 1 gram of a substance 1 degree Celsius (°C).

To calculate the amount of heat absorbed by the sample, use the formula:

Q = mcΔT

where Q is the amount of heat absorbed by the sample, m is the mass of the sample, c is the specific heat of aluminum, and ΔT is the change in temperature of the sample.

Substituting the given values into the formula, we get:

Q = 500.0 g × 0.9930 J/g·K × (90.7°C - 25.0°C)

Q = 500.0 g × 0.9930 J/g·K × 65.7 K

Q = 32,622.45 J

To convert the result to kilojoules (kJ), we divide by 1000:

Q = 32.62245 kJ

Thus, the amount of heat absorbed by the sample is 32.6 kJ.

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Answer:

The answer is B

Brainliest please!

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl  is the limiting reactant and 9 moles of NaNO3 is obtained.

9 moles of NaNO₃ is obtained.

[tex]Al(NO_3)_3 + 3NaCl-----> 3NaNO_3 + AlCl_3[/tex]

From the reaction, it is seen that the limiting reactant yields the least amount of NaNO₃.

1 mole of Al(NO₃)₃ yields 3 moles of NaNO₃

4 moles of Al(NO₃)₃ yields 4 * 3/1 = 12 moles of NaNO₃

Also,

3 moles of NaCl yields 3 moles of NaNO₃

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO₃

Hence, NaCl  is the limiting reactant and 9 moles of NaNO₃ is obtained.

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Answer:

0.29 J/g.°C

Explanation:

Step 1: Given data

Step 2: Calculate the temperature change (ΔT)

ΔT = 99.4 °C - 92.6 °C = 6.8 °C

Step 3: Calculate the specific heat of the substance (c)

We will use the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 49 J / 25.0 g × 6.8 °C = 0.29 J/g.°C

Answer:

What is

the volume of the balloon when the temperature is 310K?

Answer:

100 g

Explanation:

Primero hay que calcular el número requerido de moles de NaOH, usando la definición de molaridad:

Ahora hay que convertir 2.5 moles de NaOH en gramos, usando su masa molecular:

Se requieren 100 gramos de NaOH.

Answer: 2.20 g Pb

They gave us the masses of two reactants and asked us to determine the mass of the product. We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

           2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

Answer:

[tex]m_P=4.75x10^{19}g\ P[/tex]

Explanation:

Hello there!

In this case, according to the given atoms of phosphorous, it is possible to calculate the mass of those atoms by bearing to mind the definition of mole in terms of the Avogadro's number; which refers to the mass and amount of particles in one mole of any element as equal to the atomic mass and the Avogadro's number respectively:

[tex]1 molP=6.022x10^{23}atoms\ P=30.97gP[/tex]

Which is used to obtain the required mass of P:

[tex]m_P=9.23x10^{41}atoms\ P*\frac{30.97g P}{6.022x10^{23}atoms\ P}\\\\m_P=4.75x10^{19}g\ P[/tex]

Regards!

Answer:

a) Superficial fluid

b)  5.9*10^-2 atm

c) Gas

d) Liquid

e) Solid

Explanation:

a)  At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a superficial fluid because liquid and gases does not exit at temperature and pressure greater than  405.5 K and  111.5 atm

b) NH3 does not exist as a liquid at pressures below  5.9*10^-2 atm , That is below the triple point there is existence of liquid

c)  NH3 is a Gas at 5.90×10^-2 atm and 249.5 K.

d) NH3 is a  Liquid  at 1.00 atm and 236.0 K. because pressure and temperature ( standard ) is between the given normal melting and boiling point

e) NH3 is a solid at 24.6 atm and 185.6 K . because the pressure here is more than that of triple point while the temperature is lesser

To make 200 mL of a 10% by volume solution, add 20 mL corn syrup to water.

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Answer:

I’m pretty sure it’s C

Explanation:

AP Ex