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Liono4ka [1.6K]
3 years ago
8

A sampling of 30 million particles coming from an exhaust vent contained 60 sulfur dioxide molecules. What is another way of exp

ressing this concentration of sulfur dioxide using this information?
Chemistry
1 answer:
Masteriza [31]3 years ago
4 0
I’m not sure on this one either
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1. a) Anthony and Joseph create a start up company to make batteries for electric cars. Iron(III) phosphate is needed to create
Elenna [48]

Answer:

 LR is Na₃PO₄  

Explanation:

A quick way to determine the limiting reactant in a process is to convert reactant values to moles and then divide by the respective coefficient of the balanced equation. The smaller number of the division is the limiting reactant. For the given reaction, the rxn ratio of reactants is 1:1 so only the smaller mole value gives limiting reactant. However, if the reaction is NOT 1:1 the one must divide by respective coefficient to identify the smallest value and the limiting reactant.

This problem:

                          FeCl3(aq)          +    Na3PO4(aq)   => FePO4(s) + 3 NaCl(aq)

Given:                  27.8g                          61.9g

moles:        27.8g/162.2g/mole      1.9g/163.94g/mole

                    = 0.1714 mole               = 0.0116 mole

÷ coef.         => 0.1714/1 = 0.1714      => 0.0116/1 = 0.0116

smaller value is LR => =>  =>  =>  => => LR is Na₃PO₄              

Hope this helps. Doc :-)    

3 0
3 years ago
Ammonia gas combines with excess oxygen gas to produce nitric oxide and water. How many grams of ammonia gas would have to react
Sedbober [7]

Answer:

11.6g of NH₃(g) have to react

Explanation:

For the reaction:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ

<em>4 moles of ammonia produce 905kJ</em>

Thus, if you want to produce 154kJ of energy you need:

154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-

0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>

3 0
3 years ago
How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?
docker41 [41]
1) Write the balanced chemical equation

     2HCl + Na2 CO3 ----------> 2NaCl + H2CO3

2) Write the molar ratios:

    2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

Answer: 28 mililiters of 0.1 M HCl.
7 0
3 years ago
Can someone please help me on #17? I don’t get it :(
Dmitry [639]
Given is the specific heat of water equal to 4.18 Joule per gram per *C.

This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.

Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.

To do so we have formula

Q = m C (temperature difference)

Have a look at pic for answer

5 0
3 years ago
I’m taking a test!! Please help!!
777dan777 [17]

Answer:

I am not really sure, but I think Fr.

Explanation:

8 0
3 years ago
Read 2 more answers
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