CHEMISTRY HIGH SCHOOL

Convert 205 L to ul.

Answers

Answer 1
Answer:

Answer:

2.05 x 10⁸ uL

205000000 uL

Explanation:

Step 1: Find the conversion

1 liter (L) = 1 x 10⁶ microliters (uL)

Step 2: Set up dimensional analysis

205 L((1(10)^6uL)/(1L) )

Step 3: Multiply and cancel out units

Liters and Liter cancel out.

We are left with uL

205(1 x 10⁶) = 2.05 x 10⁸ uL or 205000000 uL
Answer 2
Answer:

Answer: 205000000 microliter

Hope this helps!

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The process of splitting an atom into two nuclei with smaller masses is called _____. transmutation nuclear fission nuclear fusion chain reaction

Answers

the process of splitting an atom into two nuclei with smaller masses is called nuclear fission

Answer:  nuclear fission

Explanation:

Just did the test and got it right. :)

Which type of heat transfer can cause the butter to melt?A. Convection
B. Conduction
C. Radiation
D. Heat

Answers

Answer:

radiation

LOL but also heat

Explanation:

hope this helps

Suggest two ways to liquefy the gases.

Answers

There are two ways to liquefy a gas:

-- Increase the pressure on it.

-- Lower its temperature.

The liquefication is faster and easier if you do both.

Calculate the molarity (M) of 1.5 L solution of todium nitrate (NaNO3) containing 61.2 g of NaNO3

Answers

Answer:

\huge\boxed{\sf Molarity = 0.48\ M}

Explanation:

Given data:

Volume of solution = 1.5 L

Mass of solute = 61.2 g

Molar mass of solute:

= 23 + 14 + (16 × 3)

= 23 + 14 + 48

= 85 g/mol

Required:

Molarity = ?

Formula:

Molarity = no. of moles of solute / volume

Solution:

First, we have to find moles of solute.

We know that,

Moles = mass / molar mass

Moles = 61.2 / 85

Moles = 0.72 moles

Now, finding molarity:

Molarity = 0.72 / 1.5

Molarity = 0.48 M

\rule[225]{225}{2}

The molarity of the 1.5 L solution of sodium nitrate containing 61.2 g of NaNO3 is approximately **0.480 M**

To calculate the molarity (M) of a solution, you can use the formula:

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \]

- Volume of solution (\( V \)) = 1.5 L

- Mass of sodium nitrate = 61.2 g

- Molar mass of sodium nitrate = 85 g/mol (Na: 23 g/mol, N: 14 g/mol, O: 16 g/mol)

First, calculate the moles of sodium nitrate using its molar mass:

\[ \text{Moles of } \text{NaNO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{61.2 \, \text{g}}{85 \, \text{g/mol}} \approx 0.720 \, \text{mol} \]

Now, substitute the values into the molarity formula:

\[ \text{Molarity (M)} = \frac{0.720 \, \text{mol}}{1.5 \, \text{L}} \]Calculate:\[ \text{Molarity (M)} \approx 0.480 \, \text{M} \]

So, the molarity of the 1.5 L solution of sodium nitrate containing 61.2 g of NaNO3 is approximately **0.480 M**.

To know more about molarity refer here:
brainly.com/question/31545539#

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What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl?

Answers

3.72/30.97=0.120/0.120=1
21.28/35.45=0.600/0.120=5

PCl5

Answer:

PCI5

Explanation:

Using the atomic masses and relative abundance of the isotopes of nitrogen given below, determine the average atomic mass of nitrogen. N-14: 14.003 amu; 99.63% N-15: 15.000 amu; 0.37%

Answers

The formula for average atomic mass is :

mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....

Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
                                                 = (14.003 * 0.9963) + ( 15.000 * 0.0037)
                                                 = 13.951 + 0.056
                                                 = 14.007 a.m.u.                                                 
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