Answer:
Nearly 80 per cent of organisations in North America and Europe were victims of cyber attacks last year and nearly half of cyber attacks used malware hidden in encrypted traffic to evade detection, a report said on Wednesday.
The encryption technology that is crucial to protecting sensitive data in transit such as web transactions, emails and mobile apps, can allow malware hiding inside that encrypted traffic to pass uninspected through an organisation’s security framework.
The report by US-based security company A10 Networks in partnership with Ponemon Institute surveyed 1,023 IT and IT security practitioners in North America and Europe, highlighting the challenges these professionals face in preventing and detecting cyber attacks.
Answer:
A system of knowledge and the methods to find that knowledge is Science.
<em>Answer:</em>
- 0.052301 km have 5 significant figure
- 400 cm have 1 significant figure
- 50.0 m have 3 significant figure
- 4500.01 ml have 6 significant figure
<em>Explanation:</em>
According to rules of significant figure
0.052301 km have 5 significant figure:
- Zero to the left of the first non zero digit not significant.
- Zero between the non zero digits are significant.
<em>400 cm have 1 significant figure:</em>
- Trailing zeros are not significant in numbers without decimal points.
<em>50.0 m have 3 significant figure:</em>
- Trailing zeros are significant in numbers when there is decimal points.
<em>4500.01 ml have 6 significant figure:</em>
- Zero between the non zero digits are significant.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
