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3 years ago

At what height the acceleration due to gravity of the earth become half ?

Chemistry

1 answer:

Serggg [28]3 years ago

3 0

Answer:

Acceleration due to gravity is reduced to half its value on the earth's surface at an altitude of 2.65×106 m

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URGENT!!! What is the balanced form of this equation?

Ostrovityanka [42]

Explanation: This is a reaction of oxidation of $H_2O_2$ in the presence of acidified $KMnO_4$ . Acidified $KMnO_4$ is a strong oxidizing agent.

To balance out the $H^+$ on the reactant side, we write $H_2O$ on the product side.

Balancing out the following reaction gives us:

$2MnO_4^-+6H^++5H_2O_2\rightarrow 2Mn^{2+}+8H_2O+5O_2$

5 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

What are the 4 different types of chemical reactions? And how to remember them

Tju [1.3M]

<span>Synthesis, decomposition, single replacement and double replacement.</span>

4 0

3 years ago

Read 2 more answers

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

almond37 [142]

The powder could be acetaminophen, analgesic having chemical formula $C_{17}H_{21}NO_{4}$

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

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

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

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

Step 1: Converting the given masses into moles.

Moles of carbon = $\frac{mass}{molar \;mass}$

Moles of carbon = $\frac{67.31g}{12g/mole}$

=5.60 moles

Moles of hydrogen = $\frac{mass}{molar \;mass}$

Moles of hydrogen = $\frac{6.978 g}{1 g/mole}$

=6.978 moles

Moles of nitrogen = $\frac{mass}{molar \;mass}$

Moles of nitrogen = $\frac{4.617 g}{14 g/mole}$

=0.329 moles

Moles of oxygen = $\frac{mass}{molar \;mass}$

Moles of oxygen = $\frac{21.10 g}{16 g/mole}$

=1.31 moles

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 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is $C_{17}H_{21}NO_{4}$ .

Learn more about the empirical formula here:

brainly.com/question/14044066

#SPJ1

3 0

2 years ago

if a sample of 4.50 moles of hydrogen gas was reacted with excess nitrogen, how many moles of ammonia gas would be produced?

sergiy2304 [10]

Answer: 3.00

Explanation: got the answer from my teachers notes

8 0

3 years ago

At what height the acceleration due to gravity of the earth become half ?​

At what height the acceleration due to gravity of the earth become half ?