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andreyandreev [35.5K]
3 years ago
7

For an atom of chlorine-35, what number would you put as the SUPERSCRIPT in the atomic notation?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
6 0
<h2>Answer:</h2><h3><em>C</em><em>I</em><em>.</em><em>S</em><em>t</em><em>r</em><em>i</em><em>c</em><em>t</em><em>l</em><em>y</em><em> </em><em>s</em><em>p</em><em>e</em><em>a</em><em>k</em><em>i</em><em>n</em><em>g</em><em>,</em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>u</em><em>b</em><em>s</em><em>c</em><em>r</em><em>i</em><em>p</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>u</em><em>n</em><em>n</em><em>e</em><em>c</em><em>e</em><em>s</em><em>s</em><em>a</em><em>r</em><em>y</em><em> </em><em>,</em><em>s</em><em>i</em><em>n</em><em>c</em><em>e</em><em> </em><em>a</em><em>l</em><em>l</em><em> </em><em>a</em><em>t</em><em>o</em><em>m</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>c</em><em>h</em><em>o</em><em>l</em><em>o</em><em>r</em><em>i</em><em>n</em><em>e</em><em> </em><em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>1</em><em>7</em><em> </em><em>p</em><em>r</em><em>o</em><em>t</em><em>o</em><em>n</em><em>s</em><em>.</em><em>H</em><em>e</em><em>n</em><em>c</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>s</em><em>o</em><em>t</em><em>o</em><em>p</em><em>e</em><em> </em><em>s</em><em>y</em><em>m</em><em>b</em><em>o</em><em>l</em><em>s</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>u</em><em>s</em><em>u</em><em>a</em><em>l</em><em>l</em><em>y</em><em> </em><em>w</em><em>r</em><em>i</em><em>t</em><em>t</em><em>e</em><em>n</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em>o</em><em>u</em><em>t</em><em>h</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>u</em><em>b</em><em>s</em><em>c</em><em>r</em><em>i</em><em>p</em><em>t</em><em>.</em><em>³</em><em>⁵</em><em>C</em><em>I</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>³</em><em>⁷</em><em>C</em><em>I</em></h3>

<h2><em><u>E</u></em><em><u>X</u></em><em><u>P</u></em><em><u>L</u></em><em><u>A</u></em><em><u>N</u></em><em><u>A</u></em><em><u>T</u></em><em><u>I</u></em><em><u>O</u></em><em><u>N</u></em><em><u>:</u></em></h2><h2><em>I</em><em> </em><em>H</em><em>P</em><em>E</em><em> </em><em>I</em><em>T</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em> </em><em>T</em><em>H</em><em>A</em><em>N</em><em>K</em><em> </em><em>Y</em><em>O</em><em>U</em><em>!</em></h2>

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9474 millimeters to centimeters using scientific notation showing work
puteri [66]

Answer:

9.474 x 10^2

Explanation:

ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1

so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2

and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm

8 0
3 years ago
What is the element present for copper
amm1812

Copper is an brown-orange color which it's atomic number is 29. With high thermal and electricity conductivity with it's smooth surface.

8 0
3 years ago
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
If a penny (with the this volume .39 cm^3) was made out of pure copper, what should its mass be?
blondinia [14]

Answer:

  3.49 g

Explanation:

The mass is the product of volume and density:

  (8.96 g/cm³)(0.39 cm³) ≈ 3.49 g

The mass of a pure-copper penny would be 3.49 g.

5 0
3 years ago
What would be the temperature change if 3.0g of water absorbed 15 J of heat?
Black_prince [1.1K]
MARK AS BRAINLIEST! :)))

3 0
3 years ago
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