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2 years ago

What is the chemical name of the compound Cu(NO3)2? Use the list of polyatomic ions and the periodic table to help you answer.

Chemistry

2 answers:

vovikov84 [41]2 years ago

5 0

Answer:

it is Copper(II) nitrate,

Explanation:

IgorC [24]2 years ago

3 0

C. Copper(II)Nitrate

Hope this helps :)

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How can you use the Periodic Table of Elements to help you find information about specific elements? PLZZZ HELP MEEEEEE!!!!

elena-14-01-66 [18.8K]

Answer:

Yuh imma help-........

6 0

2 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

A(n) ____ consists of atoms that have a specific number of protons in their nuclei.

Kamila [148]

Answer:

Solution i think

Explanation:

7 0

2 years ago

What’s the compound name for V2(S2O3)3

ohaa [14]

Answer: I HOPE THIS HELPS, HAVE A GREAT EARLY HALLOWEEN

Explanation:

PubChem CID: 1084

Molecular Formula: S2O3(2−) or O3S2-2

Synonyms: Thiosulphate THIOSULFATE ION sulfurothioate UNII-LLT6XV39PY Thiosulfate (S2O32-) More...

Molecular Weight: 112.13 g/mol

7 0

2 years ago

The reason an increase in substrate concentration does not increase the reaction rate indefinitely is because

scoundrel [369]

Answer:

Increasing substrate concentration also increases the rate of reaction to a certain point. Once all of the enzymes have bound, any substrate increase will have no effect on the rate of reaction, as the available enzymes will be saturated and working at their maximum rate.

3 0

2 years ago