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3 years ago

Which statement best compares Mercury and Earth?

Chemistry

2 answers:

Agata [3.3K]3 years ago

7 0

Answer:

Explanation:

A ) Correct, Mercury almost doesn't have an atmosphere, but Earth has an Atmosphere with oxygen and perfect for sustaining life.

B) Incorrect, Mercury is the closest planet to the sun and the second hottest planet (Venus is the most hottest)

C) Incorrect, Mercury is to hot to have liquid water,and Mercury doesn't even have an good atmosphere to have carbon dioxide.

D) Incorrect, Mercury and Venus are the two planets which do not have any moons.

Hope u learned something new :)

Please mark Brainliest.

ivanzaharov [21]3 years ago

6 0

Answer:

Explanation:c C B D are just wrong hope this helps god bless

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Calculate the energy, in joules, to heat two cubes (silver and copper), each with a volume of 10.0cm, from 15 C to 25 C

Julli [10]

Answer:

For Silver , The heat absorbed = 246.75 J

For Copper , The heat is = 343.42 J

Explanation:

The change in temperature is calculated by:

$\Delta T=T_{2}-T_{1}$

T2 = 25 C

T1 = 15 C

$\Delta T=25-15$

$\Delta T=10^{0}C$

The energy in Joules can be calculated using :

$q=mc\Delta T$

here , m = mass of the substance

c = the heat capacity

q = heat absorbed / released

We need to calculate the mass , In order to determine the value of "q".

Calculation for Silver :

The mass is calculated from the density of the element.

density of Silver = 10.5 g/ml (look at the table of density)

Volume of Cube = 10 cm^3 (given)

1 cm^3 = 1 mL

10 cm^3 = 10 mL

The mass can be calculated using the formula:

$mass = density\times Volume$

$mass = 10.5\times 10$

$mass = 105grams$

Insert the value of m , c, T in the equation.

for Silver the value of "c"= 0.235 J/gC (Look at the table)

$q=mc\Delta T$

$q=105\times 0.235\times 10$

$q=246.75J$

Calculation for Copper:

Again first calculate the mass of Copper.

Density of Copper = 8.92 g/ml

Volume = 10 mL

$mass = density\times Volume$

$mass = 8.92\times 10$

$mass = 89.2grams$

Insert the value of m , c, T in the equation.

for Silver the value of "c"= 0.385 J/gC (Look at the table)

$q=mc\Delta T$

$q=89.2\times 0.385\times 10$

$q=343.42J$

8 0

3 years ago

Get ready for the wave of screenshots :>

Reil [10]

Answer:

okay let me know when i can help, ill be happy too help

6 0

3 years ago

Read 2 more answers

*aluminium ,chromium manganese are all moderately reactive metals use the following information to arrange them in the correct r

ArbitrLikvidat [17]

The order:

$\mathrm{Al : \: Most \: reactive, reduces \: everyone!(here)}\\\mathrm{Mn}\\\mathrm{Cr}$

8 0

4 years ago

A student isolated 25 g of a compound following a procedure that would theoratically yield 81 g. Wht was his percentt yield?

Sati [7]

Answer: The percent yield of the compound is 30.86 %.

Explanation:

To calculate the percentage yield of a compound, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of compound = 25 g

Theoretical yield of compound = 81 g

Putting values in above equation, we get:

$\%\text{ yield of compound}=\frac{25g}{81g}\times 100\\\\\% \text{yield of compound}=30.86\%$

Hence, the percent yield of the compound is 30.86 %.

3 0

4 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago