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notka56 [123]
3 years ago
13

When a bond is formed, energy is _ the atoms

Chemistry
1 answer:
Otrada [13]3 years ago
5 0
"released from" the atoms
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Can yall plz help me it’s a huge major grade and i don’t know how to do this
fomenos

Answer:

1. 48 mols

2. 0.2 M

5. 1.25 L

Explanation:

Molarity= mols divided by liters

Hope this helps not sure about 3 and 4

3 0
4 years ago
What is one chemical property of water?
Setler [38]
One chemical property of water is that it can melt many substances.
3 0
3 years ago
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Is the law of conservative mass observed in this equation CaCO3 + 2HCI -->CaCI2 +H2O + CO2
pychu [463]

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:

  • Ca ⇒ 1 on the left, 1 on the right.
  • C ⇒ 1 on the left, 1 on the right.
  • O ⇒ 3 on the left, 3 on the right.
  • H ⇒ 2 on the left, 2 on the right.
  • Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

8 0
3 years ago
600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c
Ludmilka [50]

Answer:

k=1.20x10^{-3} s^{-1}

Explanation:

For a first order reaction the rate law is:

v=\frac{-d[A]}{[A]}=k[A]

Integranting both sides of the equation we get:

\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

ln\frac{[A]}{[A]_{0} } =-kt

[A]=[A]_{0}e^{-kt}

ln[A]=ln[A]_{0} -kt

k=\frac{ln[A]_{0}-ln[A]}{t}

therefore k is

k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}

8 0
4 years ago
Can someone help me with this?
amm1812

Answer:

dnaq tester

Explanation:

5 0
4 years ago
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