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3 years ago

One city is at a higher latitude than another city. How does this make the two climates different?

Chemistry

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

The higher latitude may have a colder climate than the lower latitude

Explanation:

For example, the higher the latitude of a given place (the farther away it is from the equator), the sharper the angle of the sun's rays that reach it, meaning that the rays of the sun are spread across a broader area. Therefore, higher latitudes receive less heat than lower latitude areas nearer the equator.

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Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% Ti, 6 wt% Al, and 4 wt% V. [Hin

Kamila [148]

Answer:

ρ = 4407.03 kg/m³

Explanation:

The Density of a metal alloy is given by the following equation:

1/ρ = m₁/ρ₁ + m₂/ρ₂ + m₃/ρ₃

where,

ρ = density of allow = ?

ρ₁ = density of Titanium (Ti) = 4540 kg/m³

ρ₂ = density of Aluminum (Al) = 2710 kg/m³

ρ₃ = density of Vanadium (V) = 6110 kg/m³

m₁ = mass fraction of Titanium (Ti) = 90% = 0.9

m₂ = mass fraction of Aluminum (Ti) = 6% = 0.06

m₁ = mass fraction of Vanadium (V) = 4% = 0.04

Therefore,

1/ρ = 0.9/(4540 kg/m³) + (0.06)/(2710 kg/m³) + (0.04)/(6110 kg/m³)

1/ρ = 1.9823 x 10⁺⁴ m³/kg + 0.2214 x 10⁻⁴ m³/kg + 0.0654 x 10⁻⁴ m³/kg

1/ρ = 2.2691 x 10⁻⁴ m³/kg

ρ = 1/(2.2691 x 10⁻⁴ m³/kg)

ρ = 4407.03 kg/m³

5 0

2 years ago

True or false: scientists make observations very carefully science

Olenka [21]

True, scientists do make observations very carefully.

3 0

3 years ago

Read 2 more answers

What orbital do the transition metals finish their electron configuration in

PIT_PIT [208]

Answer:

d orbitals

Explanation:

Transition metals are generally known as d-block elements. The electronic configuration of all transition elements finish in a d-orbital weather they are first row, second row or third row transition elements. This is the thread that holds all the elements of the transition series together.

This is why elements of the transition series are generally called the d-block elements.

5 0

3 years ago

i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i

9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

$C_8H_{18}+O_2\to CO_2+H_2O$

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

$C_8H_{18}+O_2\to8CO_2+H_2O$

Now, we balance the hydrogen:

$C_8H_{18}+O_2\to8CO_2+9H_2O$

And in the end, the oxygen:

$C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O$

We can multiply all coefficients by 2 to get integer ones:

$2C_8H_{18}+25O_2\to16CO_2+18H_2O$

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

Now, we substract the rectants from the produtcs:

$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

$\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ$

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

$Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ$

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, m.

First, we have to heat the ice to 0 °C, so:

$\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}$

Then, we need to melt all this mass, so we use the latent heat now:

$Q_2=n\cdot6.03kJ/mol$

Converting mass to number of moles of water we have:

$\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}$

So:

$Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g$

Adding them, we have a total heat of:

$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

$\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}$

Since we have a total of 20kg of ice, we can clculate the percent using it:

$P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%$

5 0

10 months ago

Which sentence describes an example of sublimation?

Strike441 [17]

An example of sublimation is when a dry ice changes to carbon dioxide when kept in an open container. Sublimation is a change from solid phase to gas phase without passing through the liquid state. In this example, it is clear that the dry ice is solid form and it evaporates as gas without passing through the liquid state.

5 0

3 years ago