Answer:
d. Pergoldic acid
Explanation:
chlorate = ClO₃⁻
Goldate GoO₃⁻
On this basis we can infer that HClO₄ and HGoO₄ will be similar .
HClO₄ is called perchloric acid , hence HGoO₄ will be called pergoldic acid.
Answer:
<em>The correct option is B) a bridge to connect two landmasses over a stretch of water.</em>
Explanation:
A civil engineer is a person who deals with the designing and construction of infrastructure projects like the construction of roads, buildings, bridges, airports, tunnels etc. Civil engineering is one of the most oldest disciplines in engineering and it is broken into many sub-disciplines. A civil engineer might work for a private or a government-based organization. Hence, among the options mentioned in the question, the construction of a bridge is the most likely function of a civil engineer.
Explanation:
The chemical reaction given in the question is as follows -
MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)
NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)
As we know , the value for reduction potential are -
Mn²⁺ = + 1.51 V
NO₃⁻ = +0.96 V
From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .
Hence ,
NO₃⁻ can not oxidize Mn²⁺ .
“About 300 kilometers across have irregular shapes because their internal gravity is not strong enough to compress the rock into a spherical shape” so I’m guessing it’s false ?
It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.
Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.
Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.
Using the formula;
k = 0.693/t1/2
k = 0.693/21 hours = 0.033 hr-1
Using;
N=Noe^-kt
N = mass of radioactive sample at time t
No = mass of radioactive sample initially present
k = decay constant
t = time taken
Substituting values;
2.5/320= e^- 0.033 t
0.0078 = e^- 0.033 t
ln (0.0078) = 0.033 t
t = ln (0.0078)/-0.033
t = 147 hours
Learn more: brainly.com/question/6111443