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3 years ago

What are the end products of respiration

1 answer:

4 0

Respiration:
Glucose + oxygen ----> Carbon dioxide + water (+energy)
C6H12O6 + 6O2 ----> 6CO2 + 6H2O (+energy)

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How many sulfur atoms are in 3.7 mol of SO2? × 10 S atoms

Allisa [31]

Answer: 22.3 *10^23 S atoms

Explanation:

8 0

3 years ago

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

Calcium oxide (cao) can react with carbon dioxide (co2) in a synthesis reaction. Select the most likely products of this reactio

Triss [41]

Answer: Option (b) is the correct answer.

Explanation:

A synthesis reaction is the reaction in which two compounds or atoms combine together to result in the formation of a single compound.

For example, $CaO + CO_{2} \rightarrow CaCO_{3}$

Thus, here CaO and $CO_{2}$ combine together to result in the formation of calcium carbonate compound. Therefore, it is a synthesis reaction.

Therefore, we can conclude that the most likely products of Cao + co2 → CaCO3 reaction.

4 0

4 years ago

Read 2 more answers

Calculate how much heat is absorbed by a sample that weighs 12 kilograms, has a specific heat of 0.231 kg/CJ, and is heated from

Andrei [34K]

Answer:

97 J

Explanation:

Step 1: Given data

Mass of the sample (m): 12 kg

Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)

Initial temperature: 45 K

Final temperature: 80 K

Step 2: Calculate the temperature change

ΔT = 80 K - 45 K = 35 K

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J

4 0

3 years ago

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

4 years ago