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2 years ago

19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Chemistry

1 answer:

Alecsey [184]2 years ago

4 0

Answer:

$m_{H_2O}=3.384gH_2O$

Explanation:

Hello,

In this case, the chemical reaction is:

$Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O$

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

$n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O$

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

$m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O$

Regards.

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2 years ago

Determine the number of moles of H in each sample

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C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole

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<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>

1 mole of N₂H₄ contains 4 moles of H

Therefore,

0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H

<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>

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34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H

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1 year ago

The SN1 reaction yields Entry field with incorrect answer two (number) products. This is because ______________________. Entry f

vagabundo [1.1K]

Answer:

Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule

Explanation:

An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.

However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.

The attacking nucleophile in this case may be water or CH3OH

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3 years ago

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

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Explanation:

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