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2 years ago

A magnet is an example of a

Chemistry

1 answer:

statuscvo [17]2 years ago

7 0

The answer is distance force

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Please help me I promise I will give brainliest!

iragen [17]

Answer:

1. a b c

2. a

3. c

4. west to east

3 0

3 years ago

How does nuclear reactor produce electricty

taurus [48]

With a process called Fission. This generates heat to produce steam.

7 0

3 years ago

what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg

zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

Step 1: Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

Step 2: Calculate the number of moles

PV = nRT

⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

Step 3: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

Step 4: Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F = 3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

Step 5: Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0

3 years ago

Whose main job is to help a plant cell make proteins

hoa [83]

Answer:

Water

Explanation:

Because water is very important

7 0

2 years ago

Read 2 more answers

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago