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mylen [45]
3 years ago
11

Please help with my chemistry homework! : Fuel burns in a car engine. Describe how oxides of nitrogen are produced in a car engi

ne. I stated: -very high temperatures in the car engine create a reaction of oxygen and nitrogen. But I'm unsure if this is correct and would award me the full 2 marks?
Chemistry
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

Your statement is right.

Explanation:

When fuels are burned in vehicle engines, high temperatures are produced. At this high temperature, nitrogen and oxygen from the air combine with each other to produce nitrogen monoxide (NO). When this nitrogen monoxide is released in the air from vehicle exhaust systems, it combines with oxygen present in the air to form nitrogen dioxide (NO2).

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Help help help !!!!!!
11111nata11111 [884]

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V

4 0
3 years ago
Answer and work for this problem
MArishka [77]
We can write the balanced equation for the synthesis reaction as 
     H2(g) + Cl2(g) → 2HCl(g)

We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
     mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) * 
                           (2.02 g H2 / 1 mol H2)                        
                        = 4.056 g H2

We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
     mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
                             (70.91 g Cl2 / 1 mol Cl2)
                          = 142.4 g Cl2 

Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
6 0
3 years ago
In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
Mashutka [201]
Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

<span><span /><span>Balanced chemical equation :

</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
      g NaBr  -------------------> 311 g Pb(NO3)2

331.20  g  =   2*103*311

331.20 g = 64066

mass ( NaBr ) =  64066 / 331.20

mass ( naBr)  = 193,43 g of NaBr

hope this helps!.


</span>
5 0
3 years ago
4NH3 + 3O2 --&gt; 2N2 + 6H2O
larisa86 [58]

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

 <span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

<span>2.62 mol O2</span><span> (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>

8 0
3 years ago
Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution
dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

  • 1. we calculate the moles of NaOH

\tt mol=\dfrac{mass}{MW}

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

\tt mol=\dfrac{40~g}{40~g/mol}=1~mol

  • 2. Molarity(M)

\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16

5 0
3 years ago
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