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Tpy6a [65]
3 years ago
5

 what is 19.8-17.1

Chemistry
2 answers:
Ostrovityanka [42]3 years ago
7 0

Answer: it should be 2.8

Explanation:

irina1246 [14]3 years ago
7 0
It’s 2.7 although isn’t that math, not chemistry
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1.
xenn [34]
The answer is OH.

Hope this helps!
4 0
3 years ago
Water is wet. true or false? Why?
Vikentia [17]

Answer:

True water is wet because when something is wet in this case it's water it has water on it at a molecular level. Water molecules are bonded on top of each other so it's wet.

5 0
3 years ago
What is the mass of 1.2 x 1023 atoms of arsenic?
Gre4nikov [31]

Answer:

14.93 g

Explanation:

First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

  • 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As

Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

  • 0.199 mol * 74.92 g/mol = 14.93 g

Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.

6 0
3 years ago
(Brainliest) Please Help! Please do not give me a random, gibberish answer or I will report you and give you a low rating theref
nataly862011 [7]

Answer:

3.10g

Explanation:

Please see the attached picture for the full solution.

3 0
3 years ago
sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction
prisoha [69]

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

3 0
3 years ago
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