What is the Protist kingdom made up of?

BRAINLIEST AND 20 POINTS!!!

iogann1982 [59]

Answer:

a

Explanation:

4 0

3 years ago

N2+3H2---> 2NH3

irakobra [83]

Hello!

A) How many grams of hydrogen are necessary to react completely with 50.0g of nitrogen in the above reaction?

The balanced chemical reaction is the following one:

N₂(g) + 3H₂(g) → 2NH₃(g)

To calculate the amount of hydrogen necessary we will use the following conversion factor to go from grams of nitrogen to grams of hydrogen:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{3 moles H_2}{1 mol N_2}* \frac{2,016 g H_2}{1 mol H_2}=10,7950 g H_2$

So, 10,7950 grams of H₂ are required to react with 50 g of nitrogen

B) How many grams of ammonia are produced in the reaction from the previous problem?

The balanced chemical reaction again is the following one:

N₂ + 3H₂ → 2NH₃

To calculate the amount of ammonia produced we will use the following conversion factor to go from grams of nitrogen to grams of ammonia:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{2 moles NH_3}{1 mol N_2}* \frac{17,031 g NH_3}{1 mol NH_3}=60,7968 g NH_3$

So, 60,7968 grams of NH₃ are produced from 50 g of nitrogen

C) How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of barium chloride?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Silver Chloride produced we will use the following conversion factor to go from grams of Silver Nitrate to grams of Silver Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{2 moles AgCl}{2 moles AgNO_3}* \frac{143,32 g AgCl}{1 mol AgCl}=4,2185 g AgCl$

So, 4,219 g of AgCl are produced from the reaction of 5 g of AgNO₃

D) How much barium chloride is necessary to react with the silver nitrate in the previous problem?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Barium Chloride necessary we will use the following conversion factor to go from grams of Silver Nitrate to grams of Barium Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{1 mol BaCl_2}{2 moles AgNO_3}* \frac{208,23 g BaCl_2}{1 mol BaCl_2}=3,08 g BaCl_2$

So, 3,08 g of BaCl₂ are produced from the reaction of 5 g of AgNO₃

Have a nice day!

4 0

3 years ago

The most prominent topographical feature of the ocean discovered in the 1960's was _____.

Juli2301 [7.4K]

The mountain ridge which circles the earth.

6 0

3 years ago

Complete the following radioactive decay problem

Lana71 [14]

Answer:

have a nice day everyone :p

3 0

4 years ago

A sample of brass weighing 1.203g was analyzed. The zinc in the alloy was reacted with 35.123g of HCl in excess, according to th

Sphinxa [80]

The number of moles of reactants and products can be used to obtain the mass of products formed. The results obtained are as follows;

The mass of hydrogen produced is 0.0174 grams

The mass of zinc produced is 0.5688 grams

The percentage by mass of zinc in the brass alloy is 47.3 %

The question says that HCl was in excess. It means that after the reaction, we will have unreacted HCl(aq) and ZnCl2(aq) in the remaining solution.

Hence;

Mass of ZnCl2(aq) formed = 36.309g - 35.123g = 1.186g

Number of moles ZnCl2(aq) formed = $\frac{mass of ZnCl2(aq) formed}{molar mass of ZnCl2}$

$\frac{1.186g}{136.286 g/mol}$ = 0.0087 moles

From the reaction equation;

$Zn (s) + 2HCl(aq) - > H2(g) + ZnCl2 (aq)$

To obtain the amount of Zn reacted

1 mole of Zn yields 1 mole of ZnCl2

x moles of Zn yields 0.0087 moles of ZnCl2

x = 0.0087 moles of Zn

a) 1 mole of Zn yields 1 mole of H2

0.0087 moles of Zn yields 0.0087 moles of H2

Mass of H2 = 0.0087 moles of H2 * 2 g/mol = 0.0174 grams

b)

Mass of Zn reacted = 0.0087 moles of Zn * 65.38 grams/mole = 0.5688 grams

c)

Percentage of Zn in the alloy = 0.5688 grams/1.203g * 100 = 47.3 %

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6 0

3 years ago