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marishachu [46]

3 years ago

12

If you have a voltage of 120 volts entering your house and a light bulb with 0.625 amperes flowing through it, how many watts is

your light bulb?
Plzzzz help right answer gets a Brainly!

2 answers:

Degger [83]3 years ago

5 0

Answer:

75 watts

Explanation:

Luba_88 [7]3 years ago

4 0

Answer:

75 watts

Explanation:

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STALIN [3.7K]

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4 years ago

Scientific Notation and Significant Figures Problem:

solong [7]

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3 years ago

A chemist prepares a solution of calcium bromide by weighing out 0.607g of calcium bromide into a 450ml volumetric flask and fil

Phoenix [80]

Answer:

0.00676 M

Explanation:

A chemist prepares a solution of calcium bromide by weighing out 0.607g of calcium bromide into a 450ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Mass of calcium bromide (solute): 0.607 g

Volume of solution: 450 mL

Step 2: Calculate the moles corresponding to 0.607 g of calcium bromide

The molar mass of CaBr₂ is 199.89 g/mol.

0.607 g × 1 mol/199.89 g = 0.00304 mol

Step 3: Convert the volume of solution to liters

We will use the conversion factor 1 L = 1000 mL.

450 mL × 1 L/1000 mL = 0.450 L

Step 4: Calculate the molar concentration of calcium bromide

The molarity of the solution is:

M = moles of solute / liters of solution

M = 0.00304 mol / 0.450 L

M = 0.00676 M

5 0

3 years ago

A g[dsoflghdfkjlghskdjfhgkjhsdfgsjkhgkjfhsgkjgsdkfjhgsdjfhgsdfhgsdfjhgsdf

Setler79 [48]

Answer:

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7 0

3 years ago

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about 4.4x 10 m/s Calculate the

otez555 [7]

The question is incomplete, here is the complete question:

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about $4.4\times 10^6m/s$ . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits

<u>Answer:</u> The percentage of average speed is 41 %

<u>Explanation:</u>

We are given:

Radius of helium atom = 31 pm = $31\times 10^{-12}m$ (Conversion factor: $1m=10^{12}pm$ )

So, diameter of helium atom = $(2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m$

The diameter of the atom will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $64\times 10^{-12}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of electron = $9.1095\times 10^{-31}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s$

To calculate the percentage of average speed, we use the equation:

$\text{Percentage of the average speed}=\frac{\text{Uncertainty in velocity}}{\text{Average orbital speed}}\times 100$

We are given:

Average orbital speed = $4.4\times 10^6m/s$

Putting values in above equation, we get:

$\text{Percentage of the average speed}=\frac{1.81\times 10^6m/s}{4.4\times 10^6m/s}\times 100\\\\\text{Percentage of the average speed}=41.\%$

Hence, the percentage of average speed is 41 %

3 0

4 years ago