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3 years ago

11

A car accelerates away from the starting line at 3.6 m/s2 and has the mass of 2400 kg. What is the net force acting on the vehic

le?

1 answer:

insens350 [35]3 years ago

7 0

Answer:

F = m x a

F = 2400 kg x 3.6 m/s2

F = 8,640 N

Explanation:

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What is the difference between intermolecular forces and intramolecular forces

MrMuchimi

Answer:

Intramolecular forces are the forces that hold atoms together within a molecule. Intermolecular forces are forces that exist between molecules.

Explanation:

Examples: Intermolecular forces are categorized into dipole-dipole forces, London dispersion forces and hydrogen bonding forces.

Intramolecular forces are categorized into covalent, ionic and metal bonds

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2 years ago

Read 2 more answers

How many mole ratios can be correctly obtained from the chemical equation 2Al2O3(l) ® 4Al(s) + 3O2(g)?

scoundrel [369]

Answer:

6

Explanation:

3 0

3 years ago

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

Use the chemical equation below to determine how many moles of ammonia

anzhelika [568]

Answer:

If we assume that there will be enough Hydrogen for the reaction to occur, then there will be 8 moles of NH

Explanation:

The balanced equation will look like this:

4N2 + 4H2 -> 8NH

6 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