Why all samples of a given substance have the same intensive properties?

Which is more dense? A foam ball or a bowling ball?

STatiana [176]

A bowling ball.
definition of dense-closely compacted in substance

6 0

3 years ago

5) How many moles are in 5 Liters of H2 gas?

telo118 [61]

From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
From 5 moles to liters, just multiply 5 by 22.4. I got 112 L of H2.

4 0

3 years ago

Which of the following acids is the WEAKEST? The acid is followed by its Ka value.HF, 3.5 × 10^-4HNO2, 4.6 × 10^-4HCN, 4.9 × 10^

Mila [183]

Answer:

HCN

Explanation:

There are several factors that can tell us when an acid is stronger than another, which are the following:

1. The Polarity of the X - H Bond

2. Size of the X atom.

3. Charge on the acid.

4. Oxidation state of the central atom

5. Values of Ka and pKa.

From all of this factors, we can see that the exercise is already giving us values of Ka, and also we have different types of acid, not only with the form H - X

So, we will base the force of an acid by it's pKa value.

The pKa value which is calculated with the expression:

pKa = -logKa

pKa is a value that indicates how strong is the acid you are working with. This value can vary depending on factors such the charge of the acid. I f this charge can be easily distributed in resonance structures, the compound is more acidic, and therefore the pKa value is lower.

The lower the pKa, the more acidic the compound is.

So calculating the pKa on every structure we have:

HF: pKa = -log(3.5x10^-4) = 3.46

HNO2: pKa = -log(4.6x10^-4) = 3.34

HCN: pKa = -log(4.9x10^-10) = 9.31

HCOOH: pKa = -log(1.8x10^-4) = 3.74

HClO2: -log(1.1x10^-2) = 1.96

So according to all this values, we can conclude that the weakest acid is the HCN

4 0

3 years ago

A positively charged sodium ion and a negatively charged chlorine ion are joined by

inn [45]

‘Ionic bonds’ is the answer. :D

3 0

4 years ago

Write the net ionic equation to show the formation of a solid (insoluble salt) when the following solutions are mixed. Write nor

Brrunno [24]

Answer: The net ionic equations are written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

For 1:

The chemical equation for the reaction of calcium nitrate and sodium sulfate is given as:

$Ca(NO_3)_2(aq.)+Na_2SO_4(aq.)\rightarrow CaSO_4(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ca^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$

For 2:

The chemical equation for the reaction of lead (II) nitrate and potassium chloride is given as:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)$

For 3:

The chemical equation for the reaction of calcium chloride and sodium phosphate is given as:

$3CaCl_2(aq.)+2(NH_4)_3PO_4(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4Cl(aq.)$

Ionic form of the above equation follows:

$3Ca^{2+}(aq.)+6Cl^-(aq.)+6NH_4^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4^+(aq.)+6Cl^-(aq.)$

As, ammonium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$3Ca^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)$

For 4:

The chemical equation for the reaction of barium chloride and sodium sulfate is given as:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the net ionic equations are written above.

3 0

4 years ago