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sp2606 [1]
4 years ago
7

The compound HA is an acid that is soluble in water which of the beakers in the picture shows HA behaving as a weak acid in wate

r?

Chemistry
1 answer:
lesya692 [45]4 years ago
6 0
HA is a weak acid so there should be little amount of HA and then H+ and A- in the beaker so the last one is true a weak acid never converts completely to H+ and A-
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3)<br> How many atoms are in 14 moles of cadmium?
andre [41]

Answer:

8.4322\times10^{24} atoms atoms are in 14 moles of cadmium.

Explanation:

Cadmium is a white and  silvery  metal  that is found in the curst of the earth.During the  production process of metals such as lead, copper, zinc, cadmium is extracted

Cadmium is present in some foods and is emitted when fossil fuels like as coal and oil is used, smoking cigarettes. It is used in craft glazes, metal batteries, and coatings.

We know that 1 mole of  Cadmium (Cd contains) 6.023*1023 atoms, which is Avogadro's number.

So in 14 Grams of Cadmium, the number of atoms present is,

=>14×number atoms present in one cadmium atom

=>14\times 6.023\times10^{23}

=>8.4322\times10^{24} atoms

5 0
4 years ago
You have 200 g of a substance with a molar mass of 150 g/mol. How many moles of the substance do you have? 0.75 mol 1.00 mol 1.3
tigry1 [53]

Answer:

Explanation:

200 g/150 g/mol = 1.33 mol

7 0
3 years ago
As frequency of waves increases, wavelength __________. Question 21 options: decreases increases becomes faster remains constant
tensa zangetsu [6.8K]
Frequency decreases when wavelength increase
5 0
3 years ago
Read 2 more answers
What is 7.48 torr in atm
sweet-ann [11.9K]

Answer:

0.00984211

Explanation:

7 0
3 years ago
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Need help on #129. Please help!
MrRissso [65]

The percentage yield is 72.8 %.

<em>Step 1</em>. Calculate the <em>mass of Br₂</em>

Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂

<em>Step 2</em>. Calculate the <em>theoretical yield</em>

M_r:           159.81    266.69

         2Al + 3Br₂ → 2AlBr₃

Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂

Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) =  0.2586 mol AlBr₃

Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)

= 69.05 g AlCl₃

<em>Step 3</em>. Calculate the <em>percentage yield </em>

% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %

= 72.8 %

5 0
3 years ago
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