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Mazyrski [523]
3 years ago
13

Which of these space missions visited the moon several times?

Chemistry
2 answers:
swat323 years ago
4 0

Answer:

Apollo i took the test

Explanation:

sesenic [268]3 years ago
3 0

Answer:

Voyager 1

Explanation:

Voyager 1 is a space probe launched by NASA on September 5, 1977. Part of the Voyager program to study the outer Solar System, Voyager 1 was launched 16 days after its twin, Voyager 2

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What process puts water vapor into the atmosphere?
cricket20 [7]

Answer:

transpiration

Explanation:

5 0
3 years ago
Read 2 more answers
Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h
balu736 [363]

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

                             .430mol  x 12  =  5.16 g C

                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0
3 years ago
Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of the ball relati
Alborosie

Answer:

It will have more speed/more acceleration than or the ground.

Explanation:

It will lean twords the side that moves and for ex if the wagon has a level to pull a gap down the ball will go forward

5 0
3 years ago
3 L of a gas at 250 K is changed to a temperature of 300 K. What is the final volume?
ANTONII [103]

Answer:

you can simply answer vl\t1=v2/t2

3 0
3 years ago
Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles

\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

The concentration of benzoic acid is:

C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33

pH = 14 - pOH = 14 - 5.33 = 8.67

     

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles                          

The concentration of NaOH is:

C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M

Therefore, the pH is given by this excess of NaOH:         

pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

4 0
3 years ago
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