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alex41 [277]
3 years ago
10

I need help.. thanks

Chemistry
2 answers:
Elodia [21]3 years ago
7 0

Solid - Retains its shape no matter the container

Gas - Particles move freely but dont go far

Liquid - Expands to fit the volume of container

Plasma - I guess the oter thing

(THEESE ARE JUST EDUCATED GUESSES)

professor190 [17]3 years ago
3 0
The answer is gas liquid solids
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What is the name of Cr3P
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Answer:

Chromium (I) phosphide.

Explanation:

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3 years ago
At what distance from Earth are weather maps drawn?
-Dominant- [34]

Answer:

10^5 meters

Explanation:

4 0
4 years ago
If you have a graduated cylinder, containing 15.5 mL and this volume changes to 95.2 mL after a metal with a mass of 7.95g is dr
xxTIMURxx [149]

Answer: The density of the object will be 9.97\times 10^{-2} g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of object = 7.95 grams

Volume of water displaced = volume of the object = (95.2 - 15.5) ml =79.7 ml

Putting in the values we get:

Density=\frac{7.95g}{79.7ml}=0.0997g/ml

Thus density of the object will be 9.97\times 10^{-2} g/ml

4 0
4 years ago
4.93x10^25x(6.02 x 10^23)
Dmitriy789 [7]
The answer should be
2.96786*10^49
3 0
3 years ago
Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

Lookup the relative atomic mass of \rm H, \rm S, and \rm O on a modern periodic table:

  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

6 0
3 years ago
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