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Sonja [21]
3 years ago
7

Tokyo, Japan is 8817.6 km from california. How many milimeters is this? Answer in scientific notation.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0
8.8176*10^9
Hope this helps!
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What volume of a 0.3300M solution of sodium hydroxide would br ruired to titrate 15.00 mL of 0.1500 M oalic Acid?
Ganezh [65]

Answer:

3.41 mL

Explanation:

At equivalence point  from the reaction given,

Moles of Oxalic\ Acid = 2 × Moles of NaOH

Considering

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

Given  that:

Molarity_{NaOH}=0.3300\ M

Volume_{NaOH}=?

Volume_{Oxalic\ Acid}=15.00\ mL

Molarity_{Oxalic\ Acid}=0.1500\ M

So,  

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

2\times 0.3300\times Volume_{NaOH}=0.1500\times 15.00

Volume_{NaOH}=3.41\ mL

4 0
3 years ago
Answer ASAP<br> 30 points!!<br> Why do elements and compounds qualify as pure substances?
MatroZZZ [7]

Elements are made of only one kind of atom. The molecule is made up of two or more kinds of atoms. There is no physical change that can separate the compounds into more than one kind of substance. This makes a compound a pure substance.

3 0
3 years ago
Read 2 more answers
Calculate Delta H in KJ for the following reactions using heats of formation:
lozanna [386]

Answer:

<h3>(a)</h3>

\Delta H\textdegree = -2856.8\;\text{kJ} per mole reaction.

<h3>(b)</h3>

\Delta H\textdegree = -22.3\;\text{kJ} per mole reaction.

Explanation:

What is the standard enthalpy of formation \Delta H_f\textdegree{} of a substance? \Delta H_f\textdegree{} the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, \Delta H_f\textdegree{} = 0 for the most stable allotrope of each element under standard conditions. For example, oxygen \text{O}_2 (not ozone \text{O}_3) is the most stable allotrope of oxygen. Also, under STP \text{O}_2  is a gas. Forming \text{O}_2\;(g) from itself does not involve any chemical or physical change. As a result, \Delta H_f\textdegree{} = 0 for \text{O}_2\;(g).

Look up standard enthalpy of formation \Delta H_f\textdegree{} data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam \text{H}_2\text{O} in particular) and the sign of the enthalpy changes.

  • \text{C}_2\text{H}_6\;(g): -84.0\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{CO}_2\;(g): -393.5\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{H}_2\text{O}\;{\bf (g)}: -241.8\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{PbO}\;(s): -217.9\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{PbO}_2\;(s): -276.6\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{Pb}_3\text{O}_4\;(s): -734.7\;\text{kJ}\cdot\text{mol}^{-1}

How to calculate the enthalpy change of a reaction \Delta H_\text{rxn} (or simply \Delta H from enthalpies of formation?

  • Multiply the enthalpy of formation of each product by its coefficient in the equation.
  • Find the sum of these values. Label the sum \Sigma (n\cdot \Delta_f(\text{Reactants})) to show that this value takes the coefficients into account.
  • Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
  • Find the sum of these values. Label the sum \Sigma (n\cdot \Delta_f(\text{Products})) to show that this value takes the coefficient into account.
  • Change = Final - Initial. So is the case with enthalpy changes. \Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants})).

For the first reaction:

  • \Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1};
  • \Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1};
  • \begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}.

Try these steps for the second reaction:

\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}.

6 0
3 years ago
Calculate the concentration of an anthracene solution which produced a fluorescence intensity ( I ) of 775 when the irradiance o
Kay [80]

Answer:

The concentration of an anthracene solution is  c = 3.560 *10^{-5}M

Explanation:

From the question we are told that

       The incident beam P_o is =1532

         The fluorescence intensity is  I = 775

          The length of the medium is b = 0.875 cm

         The molar extinction coefficient is  \epsilon = 9.5 *10^3 M^{-1} cm^{-1}

          The proportionality  constant k = 0.30

 

According to Lambert law the Absorbance of the anthracene solution is mathematically represented as

              A = log (I_O/I)

Where I_o =P_o

and A  is the  Absorbance

    Substituting value

                A = log( (1532)/(775))

                  =0.2960

Generally beers law can be represented mathematically as

                   A = \epsilon c l

where c is the concentration of an anthracene solution

Making c the subject of the formula

          c = \frac{A}{cl}

Substituting  0.875 cm for length = b ,

We have  

            c = \frac{0.2960}{9.5*10^{3} * 0.875}

              c = 3.560 *10^{-5}M

             

6 0
4 years ago
What quantity in moles of LiBr are in 66.1 grams of LiBr?
anygoal [31]

Molar mass of LiBr

  • 7+80=87g/mol

\\ \sf\hookrightarrow No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}

\\ \sf\hookrightarrow No\:of\:moles=\dfrac{66.1}{87}

\\ \sf\hookrightarrow No\:of\:moles=0.75mol

5 0
2 years ago
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