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4 years ago

7

How many feet are contained in .399 km

1 answer:

stiks02 [169]4 years ago

7 0

1309.0551181 feet are contained in .399 km

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According to its periodic table entry, how many electrons does nitrogen have in its valence level?

Goryan [66]

The valence level of an atom refers to the number of electrons that reside in the upper most energy level. Nitrogen has an atomic number of 7 and therefore has 7 electrons. The first energy level (1) holds 2 electrons, leaving 5 electrons to reside in the second energy level (2s and 2p). Therefore the valence of nitrogen is 5.

3 0

3 years ago

The first part of an organism's 2 part scientific name. It is always

Harrizon [31]

Answer:

Its B: Genus.

Explanation:

It has a scientific name so that's why its Genus.

5 0

3 years ago

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(l) + 15O2(g)→12CO2(g)+6H2O(l) ΔH∘=−653

Pachacha [2.7K]

Answer: 48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

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

$-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]$

$\Delta H_{C_6H_6}=48.6kJ/mol$

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

4 0

3 years ago

The following physical constants are for water, H2O.

Delicious77 [7]

Answer:

$Q\approx6.4~kJ$

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

$Q_1=m.C_s.\Delta T$

here;

mass, m = 10 g

specific heat capacity of ice, $C_s=2.09~J.g^{-1}.^{\circ}C^{-1}$

change in temperature, $\Delta T=(5-0)=5^{o}C$

$Q_1=10\times2.09\times 5$

$Q_1=104.5~J$

Amount of heat required to melt the ice at 0°C:

$Q_2=m.\Delta H_{fus}$

where, $\Delta H_{fus}=6020~J/mol$

we know that no. of moles is = (wt. in gram) $\div$ (molecular mass)

$Q_2=\frac{10}{18} \times 6020$

$Q_2=3344.44~J$

Now, the heat required to bring the water to 70°C from 0°C:

$Q_3=m.C_L.\Delta T$

specific heat of water, $C_L=4.18~J/g/^oC$

change in temperature, $\Delta T=(70-0)=70^oC$

$Q_3=10\times 4.18\times 70$

$Q_3=2926~J$

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

$Q=Q_1+Q_2+Q_3$

$Q=104.5+3344.44+2926$

$Q=6374.94~J$

$Q\approx6.4~kJ$

8 0

3 years ago

Two chemical samples were heated individually in a flame and both produced an identical orange flame. What conclusion can you dr

cestrela7 [59]

Answer: they both contain the same metal(calcium)

Explanation: calcium burns with a characteristic orange color.

6 0

4 years ago