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2 years ago

3. Jasmine pousse un objet sur une distance de

1 answer:

5 0

Jasmine pushes an object a distance of 15 m at constant speed. It produces a 450J of work on the object. What strength Jasmine does it exert on the object?

Answer:

30J

Explanation:

Given parameters:

Distance moved = 15m

Work done = 450J

Unknown:

Strength Jasmine applied = ?

Solution:

The strength Jasmine applied or exerted on the object is the force of pull that cause the motion of the object and the distance it was moved.

Now;

Work done = Force x distance

So;

450 = Force x 15

Force = $\frac{450}{15}$ = 30J

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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3 years ago

What do you call a bond between two neutral atoms that share electrons

Katyanochek1 [597]

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms

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The chemical shift of the CH3 protons in diethylether is δ = 1.16 and that of the CH2 protons is 3.36. What is the difference in

grandymaker [24]

Answer:

Where Blocal = local magnetic field between the two regions of the molecule

Blocal = (1-σ)B0

ΔBlocal = (1-σ1)B0 - (1-σ2)B0 = (σ2 - σ1)B0 = ΔσB0 ≈ ΔδB0 x 10∧-6

= (3.36-1.16) x 10∧-6 x B0 = 2.20 x 10∧-6B0

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