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nika2105 [10]
3 years ago
11

Please help me with this question it’s due in five minutes I’ll give brainliest

Chemistry
1 answer:
zysi [14]3 years ago
5 0
I believe the correct answer is A) 6
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You are on an alien planet where the names for substances and the units of measures are very unfamiliar.
neonofarm [45]

Given that in the alien planet the names of substances and the units in which they are measured in not familiar.

8 quibs of a substance called skvarnik is present.

The conversion factor between sleps and quibs is: 9 sleps = 13 quibs

Converting 8 quibs of skvarnick to sleps:

8 quibs * \frac{9 sleps}{13 quips}= 5.54sleps

5.54 when rounded to nearest tenth becomes 5.5.

So the correct answer would be 5.5



5 0
3 years ago
A graduated cylinder is filled with 60.0 ml of water. A cube, made of aluminum, is carefully dropped into the cylinder. Aluminum
katen-ka-za [31]

Answer:

65 ml

Explanation:

The aluminum will not float , so it will displace a volume of fluid equal to its volume.

 13.5 gm / 2.7 gm/ml = 5 ml

the new graduated cylinder measurement will be 60 + 5 = 65 ml

3 0
2 years ago
Which of these is an accurate source of evidence for an investigation?
Feliz [49]

Answer:

observations

Explanation:

4 0
3 years ago
Be sure to answer all parts. A person drinks four glasses of cold water (3.2 degree C) every day. The volume of each glass is 2.
Lelechka [254]

Explanation:

(a). The given data is as follows.

Volume of one glass of water = 2.2 \times 10^{2} ml = 220 ml

Volume of 4 glass of water = 220 \times 4 = 880 ml

We known that density of water is 1 g/ml.  Therefore, calculate the mass of water as follows.

       Mass of water = 880 ml \times 1 g/ml

                               = 880 gm

                               = 0.88 Kg               (as 1 kg = 1000 g)

The relation between heat energy, mass and temperature change is as follows.

            Q = mC \Delta T

\Delta T = (37 - 3.2)^{o}C = 33.8^{o}C

Putting the given values into the above formula as follows.

         Q = mC \Delta T

             = 0.88 \times 4.186 J/g^{o}C \times 33.8^{o}C

             = 124.5 kJ

Hence, the body have to supply 124.5 kJ to raise the temperature of the water to 37 degree C.

(b).      As we know that the heat of fusion of ice is 333 J/g.

So, energy required for 8.4 \times 10^{2} g or 840 g is as follows.

          333 \times 840 = 279.72 kJ

Heat capacity of water= 4.184 J/g^{o}C

Now, heat energy will be as follows.

           Q = 4.184 \times 840 g \times 37^{o}C

               = 130.03 kJ

Therefore, total heat required = (279.72 + 130.03) kJ

                                                  = 409.75 kJ

Hence, for the given situation your body should lose 409.75 kJ  of heat.

6 0
3 years ago
If you have 1.26 x 10^24 molecules, how many moles do you have?
spin [16.1K]

Answer:

E

Explanation:

Just e

7 0
3 years ago
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