Examine the statement about alchemy

The specific heat of aluminum is 0.897 J/g•°C. Which equation would you use to calculate the amount of heat needed to raise the

docker41 [41]

Q = 0.75 g x 0.897 J/g•°C x 22°C

6 0

3 years ago

Question 25———-A, YA are two isotopes of element A.

lakkis [162]

Answer:

Option C. 1

Explanation:

Step 1:

Determination of the Neutron of both isotopes. This is illustrated below.

For isotope y xA:

Mass number = y

Atomic number = x

Neutron =..?

Atomic number = proton number = x

Mass number = Proton + Neutron

y = x + Neutron

Rearrange

Neutron = y – x

For isotope (y + 1) xA:

Mass number = y + 1

Atomic number = x

Neutron =.?

Atomic number = proton number = x

Mass number = Proton + Neutron

y + 1 = x + Neutron

Rearrange

Neutron = y + 1 – x

Step 2:

Determination of the difference between the neutron number of both isotopes. This is illustrated below:

For isotope y xA:

Neutron number = y – x

For isotope (y + 1) xA:

Neutron number = y + 1 – x

Difference in neutron number

=> (y + 1 – x) – (y – x)

=> y + 1 – x – y + x

Rearrange

=> y – y + 1 – x + x

=> 1

Therefore, the difference in the neutron number of both isotopes is 1

6 0

2 years ago

Protons, neutrons, and electrons all have similar masses. explain why this statement is false.

yan [13]

It is false because electrons have no mass.

6 0

2 years ago

If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),

garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] (1)

Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻

pH = 7.40

pKa = 7.20

[A-] + [HA] = 0.10moles (2)

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0

2 years ago

Select multiple choices. Thank you

AnnZ [28]

D, systems of writing

4 0

2 years ago