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2 years ago

CHEMISTRY. PLEASE HELP!!

Chemistry

2 answers:

Zina [86]2 years ago

6 0

Answer:

B: It is positive when the bonds of the product store more energy than those of the reactants.

Explanation:

took the test

bixtya [17]2 years ago

3 0

Answer:

The answer is B on edge

Explanation:

Here are my notes on this section for anyone that needs them

Enthalpy and State Function

Bonds contain potential energy. Breaking and forming bonds involves energy. Reactants and products contain energy. Enthalpy (H) is a measure of heat and internal energy in a system.

A state function is a quantity whose change in magnitude during a process depends only on the beginning and end points the process, not the path taken between them. Enthalpy change during reaction depends only on the identity of reactants and products and their initial and finial condition

Enthalpy of Formation

enthalpy of formation (Hf) is the energy absorbed or released when a pure substance forms from elements in their standard states

Units: kJ/mol, kcal/mol

Standard state is the natural state of an element at 1 atm (atmosphere of pressure) and 25 degrees celsius. Hf for a pure element in its standard state is 0 kJ/mol.

H (hydrogen): H2(g)

N (nitrogen): N2(g)

O (oxygen): O2(g)

F (fluorine): F2(g)

Cl (chlorine): Cl2(g)

Br (bromine): Br2(l)

Hg (mercury): Hg(l)

Enthalpy of Reaction

Enthalpy of reaction (Hrxn) is energy absorbed or released during a chemical reaction

Hrxn negative: exothermic reaction

Hrxn positive: endothermic reaction

Hess's Law: Hrxn = Σ(ΔHƒ, products) − Σ(ΔHƒ, reactants)

thermochemical equation: the chemical equation that shows the state of each substance involved and the energy change involved in a reaction

Find the kJ/mol of the product and then subtract the kJ/mol of the reactants.

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This is called ____________when pressure builds up until gasses can no longer escape this substance destroys everything in its p

horsena [70]

Volcanoes can destroy each and every thing

7 0

2 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

4 0

2 years ago

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )

Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

so pH=14-9

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

4 0

3 years ago

Read 2 more answers

How many grams of NaCl are needed to prepare 60 g of a 6.0% solution?

AlekseyPX

Answer:

3.6 grams of NaCl are needed

Explanation:

Percent solution are solutions whose concentrations are expressed in percentages. The amount(either weight or volume) of a solute is expressed as a percentage of the total weight or volume of solution. Percent solutions can either be expressed as weight/volume % (wt/vol % or w/v %), weight/weight % (wt/wt % or w/w %), or volume/volume % (vol/vol % or v/v %).

A 6.0% wt/wt % solution contains 6 g of solute in 100 g of solution

Therefore, a 100 g solution contains 6.0 g of solute.

60 g of 6.0% solution will contain 60 g solution * 6.0 g solute/ 100 g solution

Mass of NaCl present = 3.6 g of NaCl

3 0

3 years ago

Joseph Priestley discovered and described the chemical properties of oxygen. What type of chemist would he be considered today?

Verizon [17]

The answer would be b. inorganic because biochemists look at chemicals containing carbon.

3 0

3 years ago