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A ring costs $36 more than a bracelet. the cost of the bracelet is 4/7 the cost of the ring. find the total of the two items.

avanturin [10]

So let's use some equations to represent the data [let R= cost of ring & B= cost of bracelet]

R= B + $ 36 .... (1)

B= $\frac{4}{7}$ × R ... (2)

By using simultaneous equations to solve for B and R.
Substitute eq. (1) into eq. (2)

B = $\frac{4}{7}$ × (B + $36)

B = $\frac{4}{7}$ B + $\frac{144}{7}$

$( \frac{7}{7} - \frac{4}{7} ) B = \frac{144}{7}$

$\frac{3}{7} B = \frac{144}{7}$

⇒ B = $48

By substituting value of B into ea (1)

If R = B + $36

R = ($48) + $36

= $84

∴ <span> the total of the two items = R + B
= $84 + $48
</span> = $132

7 0

3 years ago

Convert 13.4 degrees celcius into kelvin

Alenkasestr [34]

Answer:

286.55K

Explanation:

To convert to kelvin , add 237 .15

$13.4\°C + 273.15 = 286.55K$

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

A current of 5. 68 a is passed through a Fe(NO3)2 solution. How long, in hours, would this current have to be applied to plate o

daser333 [38]

There are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In $Fe(NO_{3} )_{2}$ , the valancy of Fe is +2 .

2 moles of $e^{-}$ are required for the decomposition of 1 mole of Fe .

7. 20 g of Fe in moles = 7. 20 g /55.845 g/mol =0.12 mole

x moles of $e^{-}$ are required for the decomposition of 0.128 mole of Fe .

moles of $e^{-}$ are required = 0.256 moles

Charge on 1 mole of $e^{-}$ = 96500 C

Charge on 0.256 mole of $e^{-}$ = 24704 C

Current ( I )= Q/t

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are 1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

To learn more about iron

brainly.com/question/18500540

#SPJ4

8 0

1 year ago

element that is neither a non-metal nor a metalloid, and is characterized by thin, dark grey filaments exhibiting magnetic prope

Digiron [165]

Yes would you explain better

6 0

4 years ago