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Sergio [31]
3 years ago
14

C₆H₁₂O₆ + 6O₂→6CO₂ + 6H₂O + Energy

Chemistry
1 answer:
NeX [460]3 years ago
7 0

Answer:

C₆H₁₂O₆ + 6O₂→6CO₂ + 6H₂O + Energy=chemical equation for cellular respiration

Explanation:

hope this help

this is james not jade

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Explain how chemical change differs from a physical change
Grace [21]
A chemical change affects on the molecular level of matter, which makes it irreversible. Combustion is a pretty good exmple. Physical changes are reversible and dont alter the formula. Hope this helped!
8 0
3 years ago
Read 2 more answers
A
meriva

Answer:

Option B is correct. A nuclear alpha decay

Explanation:

Step 1

This equation is a nuclear reaction. So it can be an alpha decay or a beta decay

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation that is showed.

A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

Option B is correct. A nuclear alpha decay

6 0
3 years ago
A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution before the addition of any lioh
snow_lady [41]
Thank you for posting your question here at brainly. Below is the solution:

 <span>moles HClO4 = 0.100 L x 0.18 M = 0.018 
moles LiOH = 0.030 L x 0.27 = 0.0081 
moles H+ in excess = 0.018 - 0.0081 = 0.0099 
total volume = 0.130 L 
[H+] = 0.0099/ 0.130= 0.0762 M 
pH = 1.12</span>
3 0
3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
Answer the question in image attached <br><br> Thanks!
arsen [322]

Answer: i think the 3,4,2

Explanation:

6 0
3 years ago
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