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Amanda
5 days ago
7

Round to the place value of the underlined digit 934, 567 the 4 is underlined

Mathematics
2 answers:
Leona [12K]5 days ago
4 0
The answer is thousand's place
Svet_ta [12.2K]5 days ago
3 0
When rounding 934,567 to the value of the underlined 4, it becomes 935,000
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7 days ago
The difference of two supplementary angles is 88 degrees. Find the measures of the angles.
Zina [11949]
The values of the two supplementary angles are 89 and 1.

To arrive at this, we set the angles as A and B.

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Math 1314 lab module 2
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1 month ago
You are planning to hire a full-time electrician who will work 40 hours per week. If you plan on giving this new hire three week
AnnZ [11884]

Option b) 1960 is the total hours worked by the electrician.

Step-by-step explanation:

It is stated that a full-time electrician will put in 40 hours weekly.

Hence, we should determine the total hours he could work in a year, which consists of twelve months.

Thus, a year consists of 365 days.

To find the number of weeks in these twelve months:

Number of weeks = Total days in a year / 7 days of the week

⇒ 365 / 7

⇒ 52.14 (approximately 52 weeks)

It translates to 52 weeks within twelve months.

Out of these 52 weeks, three weeks are designated for vacation.

The total weeks the electrician will work = 52 weeks - 3 weeks

⇒ 49 weeks.

Thus, the electrician worked for 49 weeks.

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⇒ 1960 hours.

So, the total hours the electrician worked amounts to 1960 hours, which corresponds to option b).

5 0
7 days ago
What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re
lawyer [12081]

Answer:

4.0921 reflects the logarithm of the equilibrium constant.

Step-by-step explanation:

Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

Ag^+(aq) + e^-\rightarrow Ag(s); E° = 0.80 V

Iron has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the anode.

E^{o}_{cell}=Reduction potential of cathode - Reduction potential of anode

E^{o}_{cell}=E^{o}_c-E^{o}_a

=0.80 V-(-0.41 V)=1.21 V

Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

2Ag^+(aq) + 2e^-\rightarrow 2Ag(s); E° = 0.80 V

Net reaction: Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)

n = 2

To determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

\Delta G^o=-nfE^o_{cell}

and,

\Delta G^o=-RT\ln K_{eq}

Aligning these two equations yields:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 1.21 V

R = gas constant = 8.314 J/K.mol

T = reaction temperature = 25^oC=[273+25]=298K

Substituting values into the equation, we arrive at:

2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}

\ln K_{eq}=9.3478

\log K_{eq}=\frac{9.3478}{2.303}=4.0921

4.0921 represents the logarithm of the equilibrium constant.

7 0
1 month ago
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