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Roman55 [17]
2 years ago
15

During an experiment, a chemistry student finds the concentration of a solid in a liquid by dividing 3.024 g by 0.07 mL. Using s

ignificant digits, what amount should the student record in the lab report
Chemistry
1 answer:
spin [16.1K]2 years ago
4 0

Answer:

40 g/mL

Explanation: i did it and got it right

You might be interested in
 
Katyanochek1 [597]

Answer:

The answer to your question is     V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L                    Volume 2 =

Temperature 1 = T1 = 19°C               Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg           Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

                  P1V1/T1  = P2V2/T2

-Solve for V2

                  V2 = P1V1T2 / T1P2

-Substitution

                  V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

                 V2 = 1068012 / 214620

-Result

                 V2 = 4.97 l

6 0
3 years ago
a swimming pool is 50 meters long. each day a swimmer swam total of 75 lengths of the pool for practice. how many kilometers did
Brilliant_brown [7]
50x75=3750 -- metres per day
3750x15=56250 -- metres in 15 days

Divide by 1000 to convert to km
56.250km
<u>56.25km</u>
8 0
2 years ago
Question 3 (1 point)
irina1246 [14]

false

Explanation:

i took the test

4 0
2 years ago
Read 2 more answers
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Sedimentary rocks form
grin007 [14]
Hello there!

Sedimentary rocks are formed due to layers so the answer is A.

Best wishes
-HuronGirl
5 0
2 years ago
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