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2 years ago

Diana raises a 1000. N piano a distance of 5.00 m using a set of pulleys. She pulls

Chemistry

1 answer:

Alina [70]2 years ago

5 0

Answer:

a) 250 N

b) 50 N

c) 5000 J

d) 4

e) 3.33

Explanation:

Given that weight of piano ( $F_r$ ) = 1000 N, distance moved by pulley ( $d_r$ ) = 5 m and the rope used ( $d_e$ ) =20 m

a) The effort applied ( $F_e$ ) if it was an ideal machine is:

$F_e=\frac{F_rd_r}{d_e} = \frac{1000*5}{20}=250N$

b) If the actual effort = 300 N

The force to overcome friction = actual effort - $F_e$ = 300 - 250 = 50 N

c) Work output = $F_rd_r=1000*5=5000J$

d) ideal mechanical advantage (IMA) = $\frac{d_e}{d_r} =\frac{20}{5} = 4$

e) Input force = 300 N, Therefore:

actual mechanical advantage = $F_r$ / input force = 1000 / 300 = 3.33

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Predict the initial and isolated products for the reaction. The starting material is a 6 carbon chain where there is a triple bo

satela [25.4K]

Answer:

See explanation and image attached

Explanation:

This reaction is known as mercuric ion catalyzed hydration of alkynes.

The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.

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1 year ago

What element is being oxidized in the following redox reaction?

gregori [183]

Answer:

C is the element thats has been oxidized.

Explanation:

MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)

This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.

In the oxalic acid, this are the oxidation states:

H: +1

C: +3

O: -2

In the product side, in CO₂ the oxidation states are:

C: +4

O: -2

Carbon from the oxalate has increased the oxidation state, so it has been oxidized.

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2 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

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2 years ago

2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

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1 year ago

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

Answer:

For a: The equilibrium mixture contains primarily reactants.

For b: The equilibrium mixture contains primarily products.

Explanation:

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

For a:

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

For b:

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0

1 year ago