Solid waste is eliminated from the body through the esophagus agree or disagree

Molecule of dissimilar elements with a net charge remaining is called

MariettaO [177]

It is called a polyatomic ion.

Hope this helps!!!

3 0

3 years ago

Two positively charged objects (N pole) are separated by a large distance. One of the positively charged objects (N pole) is rep

Eddi Din [679]

Answer:

B

Explanation:

Applying law of electrostatic which states that like charges repel each other and unlike charges attract each other

N and S are unlike charges that turn and make the former repulsive force (due to two like charges N and N)to reduce and attractive force between N and S to increase

6 0

3 years ago

A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

elixir [45]

Answer: $\Delta H^{0}=-173.72$ kJ/mol

Explanation: Enthalpy Change is the amount of energy in a reaction - absorption or release - at a constant pressure. So, Standard Enthalpy of Formation is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

$\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}$

$Ca(OH)_{2}_{(aq)}+2HCl_{(aq)}$ → $CaCl_{2}_{(s)}+2H_{2}O_{(l)}$

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: $\Delta H^{0}=$ -1002.82 kJ/mol

Calcium chloride: $\Delta H^{0}=$ -795.8 kJ/mol

Water: $\Delta H^{0}=$ -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

$-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]$

$-17.2=-1367.46+1002.82-2\Delta H$

$2\Delta H=17.2-364.64$

$\Delta H=-173.72$

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0

2 years ago

Why do deep earthquake do less damage then shallow ones ?

Nata [24]

Shallow ones are seismic waves from deep quakes that have to travel farther to the surface, losing energy along the way. shaking is more intense from quakes that hit close to the surface.

5 0

2 years ago

PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.

Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

$\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).$

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ii. Given the dissolution of some substance

$xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)}$ ,

the Ksp, or the solubility product constant, of the preceding equation takes the general form

$K_{sp} = [B]^y [C]^z$ .

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

$K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}$ .

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iii. Presumably, what we're being asked for here is the molar solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

$1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.$

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0

3 years ago