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alexandr1967 [171]
3 years ago
10

Help please i would really appreciate it thanks

Chemistry
2 answers:
bearhunter [10]3 years ago
4 0

Answer: stop

Explanation:

lilavasa [31]3 years ago
3 0
Here the answer is it stops.
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How many moles of CO2 would be present in a gas sample of 10 L at 25.0oC and a pressure of .77 atm?
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Use PV=nRT to solve the equation. You need to solve for n (number of moles). Don’t forget to convert the temperature to kelvins by adding 25+273. Use 0.082057 for R.
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3 years ago
Popeye wants to make a dilute spinach solution for Sweetpea's bottle. How much 3.0 M spinach solution should he add to the 500.0
Lynna [10]

Answer : The volume of 3.0 M spinach solution added should be, 50 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of spinach solution.

M_2\text{ and }V_2 are the final molarity and volume of diluted spinach solution.

We are given:

M_1=3.0M\\V_1=?\\M_2=0.30M\\V_2=500.0mL

Now put all the given values in above equation, we get:

3.0M\times V_1=0.30M\times 500.0mL\\\\V_1=50mL

Hence, the volume of 3.0 M spinach solution added should be, 50 mL

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3 years ago
Which element is in group 2 and period 7 of the periodic table
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Hey
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3 years ago
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
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