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3 years ago

Use the following equation to answer the questions below:

Chemistry

1 answer:

Gala2k [10]3 years ago

3 0

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass * Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

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What is the mass, in grams, of a pure iron cube that has a volume of 3.80 cm?

Trava [24]

Answer:

2.66 g of Fe, can be obtained from the reaction

Explanation:

Let's think the reaction:

2Fe₂O₃ + 6CO → 4Fe + 6CO₂

Ratio is 2:4, so If i have x moles of iron (III) oxide, I will produce the double of moles of Fe.

Mass / Molar mass = Mol

3.80 g / 159.7 g/m = 0.0237 moles

0.0237 moles . 2 = 0.0475 moles

Molar mass Fe = 55.85 g/m

Mol . Molar mass = Mass → 0.0475 m . 55.85 g/m = 2.66 grams

Explanation:

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3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

(1) Preparatory phase

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

(2) Pay-off phase

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that BPG has a mixed anhydride and the bond at C1 is a very high energy bond. In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

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3 years ago

Select the two variables that are held constant when testing Boyle's law in a manometer.

MrRissso [65]

Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select all that apply. V2 = k/P2 V2 = P1V1/P2 ... What two variables are held constant when testing Boyle's Law in a manometer? Temperature hope this helps

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3 years ago

Read 2 more answers

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is th

kolbaska11 [484]

Answer: When using 645 L /s of O2 in a temperature and pressure of 195°C, 0.88 atm respectively, we will get 0.355Kg /s NO

Explanation:

First we review the equation that represents the oxidation process of the NH3 to NO.

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

Second we gather the information what we are going to use in our calculations.

O2 Volume Rate = 645 L /s

Pressure = 0.88 atm

Temperature = 195°C + 273 = 468K

NO molecular weight = 30.01 g/mol

Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT

Let´s keep in mind that using this equation our constant R is 0.08205Lxatm/Kxmol

PV =n RT

n= PV / RT

n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]

n= 14.781 moles /s of O2

Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .

14.781 moles /s of O2 x 4moles NO / 5 moles O2 x 30.01g NO / 1 mol NO x 1Kg NO /1000g NO = 0.355Kg /s NO

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3 years ago

Filtering air to remove dust and pollen is it chemical or physical change

Cerrena [4.2K]

Answer:

Physical Change

Explanation:

Because the filter is removeing the dust by having the air go though a thin material.

Hope it helps!

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3 years ago