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3 years ago

8

A chemist dissolved crystals of an unknown substance into water at room temperature. He found that 33 g of the substance can be

dissolved into 100ml of water. What property of the unknown substance was the chemist most likely investigating?

1 answer:

Radda [10]3 years ago

7 0

The property that the scientist was most likely investigating was the solubility of the crystals.

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The Average Speed of the orbiting space shuttle is

maxonik [38]

Explanation:

It is given that, the Average Speed of the orbiting space shuttle is 17500 miles/hour.

We need to convert the speed in kilometers/ second

We know that,

1 mile = 1.609 km

or

1 km = 0.621 miles

1 hour = 3600 seconds

$17500\ \dfrac{\text{miles}}{\text{hour}}=17500\ \dfrac{\text{miles}}{\text{h}}\times \dfrac{1\ h}{3600\ s}\\\\=17500\times \dfrac{\text{miles}}{3600\ s}$

Now cancel the miles in numerator.

$17500\times \dfrac{\text{miles}}{3600\ s}=17500\times \dfrac{\text{miles}}{3600\ s}\times \dfrac{1.609\ km}{1\ \text{miles}}\\\\=17500\times \dfrac{1.609}{3600}\ km/s\\\\=7.82\ km/s$

So, 17500 miles/hour is equal to 7.82 km/s.

8 0

3 years ago

A Lewis acid is a substance that can

katrin2010 [14]

Accept a pair of nonbonding electrons,a Lewis acid is an electron-pair acceptor. A Lewis<span> base is any </span>substance, such as the OH-<span> ion, that </span>can<span> donate a pair of nonbonding electrons. </span>A Lewis<span> base is therefore an electron-pair donor.</span>

6 0

3 years ago

1. the mass of one electron is

kupik [55]

Answer:

Mass of one electron is 9.1 × 10⁻³¹ kg

Mass of one proton is 1.673 × 10⁻²⁷ Kg

Mass of one neutron is 1.675 × 10⁻²⁷ Kg

<u>-TheUnknownScientist</u><u> 72</u>

4 0

2 years ago

Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(

sesenic [268]

Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s)
we are going to take as
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>

</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ

4 0

3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago