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Svetradugi [14.3K]
3 years ago
9

How a strontium salt emits colored light​

Chemistry
1 answer:
Bond [772]3 years ago
3 0

Answer:

When excited electrons fall to lower energy levels, they can release energy in the form of light. metal ions in the salts used in the flame tests.

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H2O has a mc021-1.jpgHvap = 40.7 kJ/mol. What is the quantity of heat that is released when 27.9 g of H2O condenses?
babymother [125]

Energy released from changing the phase of a substance from the gas phase to liquid phase can be calculated by using the specific latent heat of vaporization. The heat of fusion of water at 0 degrees Celsius is 40.7 kJ/mol. Calculation are as follows:<span> </span>


Energy = 27.9 g (1 mol / 18.02 g) x 40.7 kJ/mol


Energy = 63.09 kJ


5 0
3 years ago
Read 2 more answers
Juan observes a material in the rocks of a hilluan hammers off a piece and then examines the pieces with a hana tena He
Nata [24]

Answer:

I definitely think he mostly observed that it was clear in color.

Explanation:

5 0
3 years ago
Non-examples of air pressure
Anuta_ua [19.1K]
No pressure







Ahhhhhh (filling character req)
8 0
3 years ago
3.0 cm x 4.0 cm x 1.0 cm<br><br>[?]cm^3​
CaHeK987 [17]

Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>3</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>1</em><em>2</em><em>×</em><em>1</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em>answer is</em><em> </em><em>1</em><em>2</em><em> </em><em>cm</em><em>^</em><em>3</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

4 0
3 years ago
A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri
baherus [9]

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

3 0
2 years ago
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