Answer:

Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.
For example,
is to be written as
in engineering notation.
The given number:
0.00000009345 can be written as 
Answer upto 4 significant digits = 
Answer:
attached below
Explanation:
Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum
a) Acetone CH₃COCH₃
Attached below is the structure
b) But-2-yne (CH₃C)₂
Attached below is the structure
Answer:
The element that has been oxidized is the N
Explanation:
Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)
See all the oxidation states:
Zn²⁺ → acts with +2
In ammonia, H acts with +1 and N with -3
Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.
Zn changed from 2+ to 0. The oxidation number, has decreased.
This element has been reduced.
NO₃⁻ (aq) it's a ion, from nitric acid.
N acts with +5
O acts with -2
The global charge is -1
The N, has increased the oxidation state, so this element is the one oxidized.
<h3>Answer:</h3><h2>Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.</h2>
Answer:
3.6
Explanation:
Step 1: Given data
- Concentration of formic acid: 0.03 M
- Concentration of formate ion: 0.02 M
- Acid dissociation constant (Ka): 1.8 × 10⁻⁴
Step 2: Calculate the pH
We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.
![pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%3D%20-log%201.8%20%5Ctimes%2010%5E%7B-4%7D%20%2B%20log%20%5Cfrac%7B0.02%7D%7B0.03%7D%20%3D%203.6)