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4 years ago

Which statement is true of the motion of the ball?

1 answer:

3 0

Your answer would be answer choice D. The ball is accelerating as the velocity is increasing at a constant rate.

Hope this helps,
♥A.W.E.S.W.A.N.♥

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Car battery energy that makes device work

MAXImum [283]

Engine is it ok ksndjsjja

4 0

3 years ago

Read 2 more answers

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ

Ahat [919]

Answer: The Gibbs free energy of the reaction is 21.32 kJ/mol

Explanation:

The chemical equation follows:

$\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}$

$\text{[Oxaloacetate]}=0.130mM$

$[NADH]=2.0\times 10^2mM$

$\text{[Malate]}=1.37mM$

$[NAD^+]=490mM$

Putting values in above expression, we get:

$\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol$

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0

3 years ago

For a typical titration, 0. 010 m naoh is the titrant (in the buret). if the initial buret reading is 2. 45 ml, and the final bu

pickupchik [31]

16.25 ml NaOH was used for the titration.

<h3>Titration:</h3>

"The process of calculating the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point)" is the definition of titration.

A titration is a method for figuring out the concentration of an unknown solution by using a solution with a known concentration. Until the reaction is finished, the titrant (the known solution) is typically added from a buret to a known volume of the analyte (the unknown solution).

0. 010 m NaOH is the titrant (in the buret).

The initial buret reading is 2. 45 ml

The final buret reading is 18. 70 ml

The volume of NaOH = 18.70ml - 2.45 ml

= 16.25 ml

Therefore, 16.25 ml NaOH was used for the titration.

Learn more about titration here:

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5 0

1 year ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{78.10g/mol}{13g/mol}=6$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 6)}H_{(1\times 6)}=C_6H_6$

Hence, the molecular formula for the compound is $C_6H_6$

8 0

3 years ago

Determine the mass of chloride (MW = 35.45 g/mol ) in grams present in 100 mL of a 0.273 M solution of aqueous FeCl₃ (iron(III)

Naily [24]

Answer:

The mass chloride is 2.903 grams

Explanation:

Step1: calculate the moles of FeCl3

moles = molarity x volume in liters

volume in liters = 100 ml /1000 = 0.1 L

molarity = 0.273 M = 0.273 mol / L

moles = 0.273 mol /L x 0.1 L = 0.0273 moles

step 2: find the moles of Cl

since there is 3 atoms of cl in FeCl3

= 3 x 0.0273 =0.0819 moles

Step 3: calculate mass of cl

mass= moles x molar mass

=0.0819 moles x 35.45 g/mol = 2.903 g

7 0

3 years ago