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astra-53 [7]
3 years ago
6

2. Why are stain used when preparing observing cells under the

Chemistry
2 answers:
ella [17]3 years ago
8 0

Answer:

to see the cells clearly

erica [24]3 years ago
4 0

Answer:

So u dont mess it up

Explanation:

Suu whoop real right blood gang on dead homies

You might be interested in
Compare the relative strength of intermolecular forced in solids liquids and gases
DIA [1.3K]

Answer:

Intermolecular forces are weaker attractions that hold molecules or noble gas particles close together when they are in a liquid or solid form.

Explanation:

3 0
4 years ago
How many protons electrons and neutrons respectively does 29 CS have
irakobra [83]

55 \; \text{e}^{-}, 55 \; \text{p}^{+}, and 74 \; \text{n}^{0}- in case cesium-129 is what you were referring to.

Cesium has atomic number 55 as seen on a modern periodic table. A cesium atom nucleus would thus contain an equal number of protons; in addition to that, a <em>neutral</em> atom of cesium would contain the same number of electrons in its electron cloud.

For a particular isotope,

\text{Mass number} = \text{Number of protons per nuclei} + \text{Number of Neutrons per nuclei}.

This isotope of cesium has a mass number of 129 and contains 55 protons in each atomic nucleus. One cesium nuclei would therefore contain 129 - 55 = 74 neutrons.

8 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
3 years ago
Would you expect hydrogen chloride to be a gas, a liquid or a solid, at room temperature and pressure? Explain your answer
Vlada [557]
At room temperature hydrogen chloride is a colorless gas with a sharp or pungent odor. Under pressure or at temperatures below –85°C (-121°F), it is a clear liquid.
6 0
3 years ago
What are the mole fraction and the mass percent of a solution made by dissolving 0.21 g KBr in 0.355 L water? (d = 1.00 g/mL.) m
IRISSAK [1]

Answer:

Mol fraction H2O = 0.99991

Mol fraction KBr = 0.00009

mass % KBr = 0.059 %

mass % H2O = 99.941 %

Explanation:

Step 1: Data given

Mass of KBr = 0.21 grams

Molar mass KBr = 119 g/mol

Volume of water = 355 mL

Density of water = 1.00 g/mL

Molar mass water = 18.02 g/mol

Step 2: Calculate mass water

Mass water = 355 mL * 1g /mL

Mass water = 355 grams

Step 3: Calculate moles water

Moles water = mass water / molar mass water

Moles water = 355 grams / 18.02 g/mol

Moles water = 19.7 moles

Step 4: Calculate moles KBr

Moles KBr = 0.21 grams / 119 g/mol

Moles KBr = 0.00176 moles

Step 5: Calculate total moles

Total moles = 19.7 moles + 0.00176 moles

Total moles = 19.70176 moles

Step 6: Calculate mol fraction

Mol fraction H2O = 19.7 moles / 19.70176 moles

Mol fraction H2O = 0.99991

Step 7: Calculate mol fraction KBr

Mol fraction KBr = 0.00176 / 19.70176

Mol fraction KBr = 0.00009

Step 6: Calculate mass %

mass % KBR = (0.21 grams / (0.21 + 355) grams) *100%

mass % KBr = 0.059 %

mass % H2O = (355 grams / 355.21 grams) *100%

mass % H2O = 99.941 %

8 0
3 years ago
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