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2 years ago

What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?

1 answer:

5 0

First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER

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An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

object distance from mirror is

$d_o = -50 cm$

Focal length of the mirror is given as

$f = 40 cm$

now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

$\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}$

$d_i = 22.22 cm$

6 0

3 years ago

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Sidana [21]

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2 years ago

A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output

Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

$P=5\ V\times 1.97\ A$

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 0

2 years ago

How many Joules of potential

s2008m [1.1K]

Answer:

40 j, 80j.

Explanation:

P.E= mgh. G=10 m/s².

For 4m, P.E=1*10*4=40 joules.

For 8m, P.E=1*10*8=80 joules.

4 0

3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$