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3 years ago

How many different states of matter exist?

Chemistry

2 answers:

STatiana [176]3 years ago

7 0

There are 4 states of matter. Gas, Liquid, Solid, & Plasma

Natalka [10]3 years ago

7 0

1 Solid

2 Liquid

3 Gas

4 Plasma

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You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

bonufazy [111]

Answer:

$\frac{[F^{-}]}{[HF]}$ is larger

Explanation:

$pK_{a}=-logK_{a}$ , where $K_{a}$ is the acid dissociation constant.

For a monoprotic acid e.g. HA, $K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$ and $\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}$

So, clearly, higher the $K_{a}$ value , lower will the the $pK_{a}$

In this mixture, at equilibrium, $[H^{+}]$ will be constant.

$K_{a}$ of HF is grater than $K_{a}$ of HCN

Hence, $(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})$

So, $\frac{[F^{-}]}{[HF]}$ is larger

5 0

3 years ago

When you receive a loan, the money the lender gives you is called the ____________. A. Interest B. Line of credit C. Principal D

Reika [66]

C. Principal !!!!!!!!!

8 0

4 years ago

Read 2 more answers

Which of the following equations does not demonstrate the law of conservation of mass?

enot [183]

The third option does not obey the law of conservation of mass.

Option 3.

Explanation:

The law of conservation of mass states that the sum of the masses of reactants should be equal to the sum of the masses of the products.

For example, if we consider the first option to verify if it obeys law of conservation of mass or not, 2 Na + Cl₂ → 2 NaCl

So one way to verify it is to find the mass of Na, then multiply it with 2, and then add this with 2 times of mass of chlorine. So this sum should be equal to the 2 times mass of NaCl. But it is somewhat lengthy.

Another way to easily determine this is to check if the elements are present equally in both sides. Such as, in reactant side and product side 2 atoms of Na is present . Similarly, the Cl atoms are also present in equal number in both reactant and product side. Thus this obeyed the law of conservation of mass.

Like this, if we see the second option, there also 1 atom of Na is present in reactant and product side and 2 molecules of H is present in reactant and product side, 1 oxygen is present in reactant and product side and 1 Cl is present in reactant and product side. So it also obeys the law of conservation of mass.

But in the third option, P₄ + 5 O₂→ 2 P₄O₁₀, here, there is 4 atoms of P in reactant side but in product side there is (4*2) = 8 atoms of P. Similarly, the number of atoms of oxygen in reactants and product side is also not same. So the third option does not obey the law of conservation of mass.

The fourth option also obeys the law of conservation of mass as the number of atoms of each element is same in both the product and reactant side.

Thus, the third option does not obey the law of conservation of mass.

5 0

3 years ago

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

Best regards!

4 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago