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SVETLANKA909090 [29]
2 years ago
13

Your local grocery store

Chemistry
1 answer:
umka2103 [35]2 years ago
3 0

The total number of matchbooks required would 5,000 be and the cost would be $144

<h3>Mathematical ratios</h3>

50 matchbooks = $1.44

Each matchbook = 0.005 grams red phosphorus

25 grams of red phosphorus is needed:

                                 25/0.005 = 5,000 matchbooks

5,000 matchbooks will cost:

                         5000/50 x 1.44 = $144

More on mathematical ratios can be found here: brainly.com/question/20387079

#SPJ1

       

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Nitrogen combines with oxygen to form nitrogen monoxide (no) and nitrogen dioxide (no2). a sample of no consists of 1.14 g of ox
Sever21 [200]
2N2 + 3O2 ---> 2NO + 2NO2  
You should do it based on moles and not grams. 
1.14 g O = 0.071 moles O 
1 g N = 0.071 moles N  
So in NO2 you need 2 moles O for each mole of N 
1 g N = 0.071 moles, so you need 0.071 x 2 moles of O = 0.0.142 moles O 
0.142 moles O x 16 g/mol = 2.27 grams of O. So, you are actually correct because your answer is 2.28 grams. I just prefer to work it out in moles so it makes perfect chemical sense.
4 0
3 years ago
Read 2 more answers
Estimate the freezing and normal boiling points of 0.25 m aqueous solutions of nh4no3 nicl3 al2so43
maks197457 [2]
The items are answered below and are numbered separately for each compound. 

The freezing point of impure solution is calculated through the equation,
     Tf = Tfw - (Kf)(m)

where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.

1. NH4NO3
    Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C

2. NiCl3 
    Tf = 0°C - (1.86°C/m)(0.25 M)(4) = -1.86°C

3. Al2(SO4)3
      Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C

For boiling points, 
    Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.

1. NH4NO3
     Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C

2. NiCl3
     Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C

3. Al2(SO4)3
    Tb = 100°C + (0.51°C/m)(0.25 M)(5) = 100.6375°C
7 0
4 years ago
Over a period of years, the average rainfall in an area has decreased, causing environmental changes. Most likely to survive the
zalisa [80]

ans D) with a great amount of genetic variation

3 0
3 years ago
Read 2 more answers
Pls answer quickkk :)))
postnew [5]

Answer:

c turns back when iodine is added result indicated a chemical reaction

4 0
3 years ago
A 25.00 mL sample of of a sulfurous acid solution is completely neutralized using 17.12 mL of 0.13 M sodium hydroxide. What is t
Kryger [21]

Answer:

0.0445 M is the initial concentration of sulfurous acid.

Explanation:

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=25.00mL\\n_2=1\\M_2=0.13 M\\V_2=17.12 mL

Putting values in above equation, we get:

2\times M_1\times 25.00 mL=1\times 0.13 M\times 17.12 mL

x=\frac{1\times 0.13 M\times 17.12 mL}{2\times 25.00}=0.0445 M

0.0445 M is the initial concentration of sulfurous acid.

8 0
3 years ago
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