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faltersainse [42]

3 years ago

10

Mara and ivy are asked to identify two minerals in science class. To do this, they decided to study the physical properties of t

he minerals in the following ways.
1 examine how they freeze
2 determine how they break
3 rub them on a white tile plate . What is the best method for the girls to study the physical properties and

2 answers:

scZoUnD [109]3 years ago

8 0

Answer:

2 only is a pretty accurate answer the others dont make sense to me.

Explanation:

djverab [1.8K]3 years ago

8 0

Answer: determine how they break

Explanation: :)

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2 years ago

In which state of matter do particles have the least amount of energy

kogti [31]

Answer:

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3 years ago

What are the formulas of the acids: <br>AsO4<br>CIO4 ( l )<br>S ( ll )<br>F ( l )<br>PO4 (lll)

SVETLANKA909090 [29]

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Sorry I don’t know all of them, good luck though! :)

4 0

3 years ago

_______ is the formation of alcohol from sugar.

Katarina [22]

The best and most correct answer among the choices provided by the question is the fourth choice "alcoholic fermentation"

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I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

3 years ago

Read 2 more answers

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago