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Nimfa-mama [501]
2 years ago
11

What term is best described as the rate of a chemical reaction at any instant in time?

Chemistry
1 answer:
Fantom [35]2 years ago
5 0

Answer:

instantaneous rate would be the term.

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1
balandron [24]

Answer:

Judges and justices serve no fixed term — they serve until their death, retirement, or conviction by the Senate. By design, this insulates them from the temporary passions of the public, and allows them to apply the law with only justice in mind, and not electoral or political concerns.

6 0
3 years ago
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Sodium (NA) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec
garri49 [273]
C= 1 Valence electron
3 0
3 years ago
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A cold object will have_ kinetic energy which means _ faster moving particles
qwelly [4]

Answer: if this is true or false, then this is false, something that is cold has slow moving particles, fast moving ones generate heat which will make it warm, this object has potential energy not kinetic.

5 0
3 years ago
2C4H10+13O2-->8CO2+10H2O Using the predicted and balanced equation, How many Liters of CO2 can be produced from 150 grams of
Anna11 [10]

Answer:  233 L of CO_2 will be produced from 150 grams of  C_4H_{10}

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}  \text{Moles of} C_4H_{10}=\frac{150g}{58g/mol}=2.59moles

The balanced chemical equation is:

2C_4H_{10}+13O_2(g)\rightarrow 8CO_2+10H_2O  

According to stoichiometry :

2 moles of C_4H_{10} produce =  8 moles of CO_2

Thus 2.59 moles of C_4H_{10} will produce=\frac{8}{2}\times 2.59=10.4moles  of CO_2  

Volume of CO_2=moles\times {\text {Molar volume}}=10.4moles\times 22.4mol/L=233L

Thus 233 L of CO_2 will be produced from 150 grams of  C_4H_{10}

8 0
3 years ago
A hypothetical covalent molecule, X–Y, has a dipole moment of 1.93 1.93 D and a bond length of 109 pm. 109 pm. Calculate the par
Gnesinka [82]

Answer:

q= 110.5 ke

Explanation:

Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.

μ = q * d

μ= Dipole moment (1.93 D)

q= partial charge on each pole

d= separation between the poles(109 pm).

e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)

So,

q= \frac{1.93}{109 * 10^{-12} } coulombs

q = \frac{1.93}{109 * 10^{-12} *  1.60217662 * 10^{-19} } e

q = 1.105 * 10⁵ e

q= 110.5 ke

4 0
3 years ago
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