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3 years ago

List the number of each type of atom on the right side of the equation pb(no3)2(aq)+2nacl(aq)→pbcl2(s)+2nano3(aq)

2 answers:

6 0

The number of each type of atom on the right side of the equation

that is

Pb(No3)2(aq) + 2 NaCl(aq) = PbCl2(s) +2NaNo3(aq)

are 1 atom of Pb
2 atoms of N
6 atoms of O
2 atoms of Na
2 atoms of Cl

Explanation
In Pb(NO3)2 there is 1 x1=1 atoms of Pb, 1x2 =2 atoms of N and 3x2 =6 atoms of O

In NaCl there is 2 x1= 2 atoms of Na and 2 x1=2 atoms of Cl

Savatey [412]3 years ago

4 0

Answer:

Pb: 1 atom

Cl: 2 atoms

Na: 2 atoms

N: 2 atoms

O: 6 atoms

Explanation:

1) RighT side of the equation: PbCl₂(s)+2NaNO₃(aq)


2) The coefficients and subscripts tells the number of atoms in each chemical formula by this convention:


i) The coefficient before the compound (chemical formula) applies to all the elements to the right (in the very chemical formula).

ii) The subscripts to the right of the element apply only to the element to which it is subscripted.

ii) If there are parenthesis the subscrpts apply to all the elements inside the parenthesis.


3) With that, let's see each compound (chemical formula):


i) Pb Cl₂ (s): there are two kind of atoms, Pb and Cl.


Pb does not have any subscriPT, which is interpreted as subscript 1, so that is 1 atom of Pb.


Cl₂ means 2 atoms of chlorine


ii) 2NaNO₃:


Na: 2 atoms (because the coefficient)

N: 2 atoms (because the coefficient)

O: 6 atoms ( 2 x 3)

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(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would

Lelu [443]

Answer:

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0

3 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)

HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)

H₂O: H₂ + 1/2O₂ → H₂O (3)

NaCl: Na + 1/2Cl₂ → NaCl (4)

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

What is another name for a homogeneous mixture?

statuscvo [17]

The answer is a solution im pretty sure <3

4 0

3 years ago

A 1.5 M solution of NaOH was made in a laboratory. If the solution made had a volume of 4.5 L, how many grams of NaOH were added

Crank

Answer:

270g

Explanation:

Given parameters:

Concentration of NaOH = 1.5M

Volume = 4.5L

Unknown

Mass of NaOH added = ?

Solution:

To solve the problem, we need to find the number of moles of the NaOH first;

Number of moles = concentration x volume

Number of moles = 1.5 x 4.5 = 6.75mol

Now;

Mass = Number of moles x molar mass

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass = 6.75 x 40 = 270g

6 0

2 years ago

Ethanol (c2h5oh) melts at -114c and boils at 78 c the enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vaporiza

vovikov84 [41]

100 kilo joules There are several phases that this problem undergoes and the final answer is the sum of all the energy used for each phase. Phase 1. Heating of solid ethanol until its melting point. Phase 2. Melting of the ethanol until it's completely liquid. Phase 3. Heating of the liquid ethanol until it reaches its boiling point. Phase 4. Boiling the ethanol until it's completely vapor. To make things more interesting, some of our constant are per gram and some others are per mole. So let's calculate how many moles of ethanol we have. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass ethanol = 2*12.0107 + 6*1.00794 + 15.999 = 46.06804 g/mol Moles ethanol = 75g / 46.06804 g/mol = 1.628026719 mol Phase 1. Use the specific heat of solid ethanol and multiply by the number of degrees we need to change by the mass we have. So 0.97 J/g*K * 75 g * (-114c - -120c) = 0.97 J/g*K * 75 g * 6K = 436.5 J Phase 2: Time to melt. Just need the moles and the enthalpy of fusion. So: 1.628026719 mol * 5.02 kJ/mol = 8.172694128 kJ Phase 3: Heat to boiling. Just like heating to melting, just a different specific heat and temperature 2.3J/g*K * 75g * (78c - -114c) = 2.3J/g*K * 75g * 192 K = 33120 J Phase 4: Boil it to vapor. Need moles and enthalpy of vaporization. So 1.628026719 mol * 38.56 kJ/mol = 62.77671027 kJ Now let's add them together: 436.5 J + 8.172694128 kJ + 33120 J + 62.77671027 kJ = 0.4365 kJ + 8.172694128 kJ + 33.120 kJ + 62.77671027 kJ =104.5059044 kJ Since the least precise datum we have is 2 significant figures, round the result to 2 significant figures, giving 100 kilo joules.

3 0

3 years ago