Given the nuclear equation below, this equation is an example of

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{100}{1000}kW=0.1 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.1 kW\times 2190 h=219 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

$Cost=\$ (219\times 0.15)=\$ 32.85$

Therefore, annual cost of incandescent light bulb is $\$ 32.85$

(b) Power of bulb is 25 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{25}{1000}kW=0.025 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.025 kW\times 2190 h=54.75 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

$Cost=\$ (54.75\times 0.15)=\$ 8.21$

Therefore, annual cost of fluorescent bulb is $\$ 8.21$ .

7 0

3 years ago

4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration

anzhelika [568]

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

$percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100$

<u><em>percentage composition of potassium iodide= 0.017%</em></u>

Finally, the percent composition of potassium iodide in this sample is 0.017%.

Learn more about percent composition:

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3 0

2 years ago

Indicate the number of unpaired electrons for following:

Viktor [21]

Ignoring the n's, there would only be one unpaired electron.

6 0

3 years ago

Three elements are represented by the letters A, B, and C. the three elements have the same number of valence electrons. Element

AfilCa [17]

Answer:
A- beryllium
B- calcium
C- magnesium

Explanation
NOTE: all element in group 2 have 2 balance electrons

First let’s start with B- number of electrons= number of protons which is equal to the atomic number. therefore, the answer is calcium as it’s atomic number is 20

C- magnesium will have three energy levels considering it has 12 electrons (2,8,2).

A- beryllium is the lightest one in group 2 as it has the atomic mass of 9.0122.

3 0

3 years ago

Are insert gases used in fluorescent

aksik [14]

Hydrofluorocarbons (HFCs), perfluorocarbons (PFCs), sulfur hexafluoride (SF6) and nitrogen trifluoride (NF3).

4 0

3 years ago