The change in its axis or orbit

A light bulb has a resistance of 96.8 Ω. What current flows through the bulb when it is connected to a 120 V source of electrica

Finger [1]

Answer:

I = 1.23 A

Explanation:

Given that,

The resistance of the lightbulb, R = 96.8 Ω

Voltage, V = 120 V

We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.

$V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{120}{96.8}\\\\I=1.23\ A$

So, the current flows through the bulb is 1.23 A.

8 0

2 years ago

Advantages and disadvantages of titration?

alisha [4.7K]

Answer:

What are the advantages of titration?

Titrimetric analysis commonly referred to as volumetric analysis offers distinct advantages over cumbersome gravimetric methods:

Speed of analysis.

Instantaneous completion of reactions.

Greater accuracy due to minimization of material loss involved in decanting, filtration, precipitation or similar operations.

Explanation:

Disadvantages

It is a destructive method often using up relatively large quantities of the substance being analysed.

It requires reactions to occur in a liquid phase, often the chemistry of interest will make this inappropriate.

It can produce significant amounts of chemical waste which has to be disposed of.

It has limited accuracy.

hope this helps Please inform me if its helpful

5 0

2 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

How could we increase the strength of a solenoid?

12345 [234]

Strength of the magnetic field can be increased by:

1) increasing the current in the coil.

2) increasing the number of coils in the solenoid; and.

3) using a soft iron core within the solenoid.

Hope this helps!

8 0

3 years ago

Question 3

stepladder [879]

This must be your work, otherwise it's against the Brainly code of conduct. If you need help finding a newspaper, go to BBC. They have plenty of relevant articles.

4 0

1 year ago