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3 years ago

What is the energy conversion that occurs in cellular respiration?

1 answer:

fomenos3 years ago

3 0

Cellular respiration is the process by which the chemicalenergy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.

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Evan has a rock. The rock has a mass of 12g and a volume of 4cm3. What is the density of the rock?

Nady [450]

Multiply then divide 12g by 4cm3 because 4cm3 equals 64cm divide by 12g or multiply which ever is in the options

3 0

4 years ago

I am in need of help!!! I need someone who is really good with Chemistry to message me asap would be best if at all possible

timofeeve [1]

Im good at it message me so ya hope this helps you

6 0

4 years ago

The empirical formula for this compound that contain 31.14%sulfur and 68.86%chlorine by mass

Kaylis [27]

The empirical formula is SCl_2.

The empirical formula (EF) is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of S to Cl.

Assume that you have 100 g of sample.

Then it contains 31.14 g S and 68.86 g Cl.

Step 1. Calculate the moles of each element

Moles of S = 31.14 g S × (1 mol S/(32.06 g S) = 0.971 30 mol S

Moles of Cl = 68.86 g Cl × (1 mol Cl/35.45 g Cl) = 1.9425 mol Cl

Step 2. Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

S:Cl = 0.971 30: 1.9425 = 1:1.9998 ≈ 1:2

Step 3: Write the empirical formula

EF = SCl_2

5 0

3 years ago

Neon is compressed from 100 kPa and 16°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

Viktor [21]

Answer:

$\Delta v=-0.952m^3/kg$

$\Delta h=0$

Explanation:

In this case, since neon could be considered as an ideal gas, the specific volumes at the first and second state are respectively:

$v_1=\frac{Rg*T_1}{p_1}=\frac{0.4119kJ/kg*K*289.15}{100kPa} =1.19m^3/kg\\$

$v_2=\frac{Rg*T_2}{p_2}=\frac{0.4119kJ/kg*K*289.15}{500kPa} =0.238m^3/kg\\$

Thus, the change in the specific volume turns out:

$\Delta v=v_2-v_1=0.238m^3/kg-1.19m^3/kg\\\Delta v=-0.952m^3/kg$

Which value has sense for the compression.

In addition, the specific enthalpy change just depend on the temperature as it is an ideal gas, therefore, since the process es isothermic:

$\Delta h=Cp\Delta T=0$

Best regards.

3 0

4 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

Brut [27]

Answer:

The molar mass of unknown gas is 145.82 g/mol.

Explanation:

Volume of oxygen gas effused under time t = 8.24 mL

Effusion rate of oxygen gas = $R=\frac{8.24 mL}{t}$

Molar mass of oxygen gas = 32 g/mol

Volume of unknown gas effused under time t = 3.86 mL

Effusion rate of unknown gas = $R'=\frac{3.86 mL}{t}$

Molar mass of unknown gas = M

Graham's Law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{R}{R'}=\sqrt{\frac{M}{32 g/mol}}$

$\frac{\frac{8.24 mL}{t}}{\frac{3.86 mL}{t}}=\sqrt{\frac{M}{32 g/mol}}$

$M=\frac{32 g/mol\times 8.24 \times 8.24}{3.86\times 3.86}=145.82 g/mol$

4 0

3 years ago