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Yuki888 [10]
3 years ago
9

Calculate the average time the car took to reach each checkpoint. Record the average time in Table D of your Student Guide. The

average time to the first quarter checkpoint is seconds. The average time to the second quarter checkpoint is seconds. The average time to the third quarter checkpoint is seconds. The average time to the finish line is seconds.
The Answers:
1.39
2.18
2.82
3.36
Chemistry
2 answers:
Schach [20]3 years ago
6 0

Answer:

1.39 2.18 2.82 3.36

Explanation:

alisha [4.7K]3 years ago
5 0

Answer:2.07

3.16

4.11

4.92

Explanation:

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4 years ago
A reaction occurs when solid X is placed into solution Y. As a result, the temperature of the new solution increases by 3°C. The
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adding more of solid X will increase the heat of the solution by another few degrees

4 0
3 years ago
Read 2 more answers
What is the equation for photosynthesis?
AlladinOne [14]

Answer: the process of photosynthesis is commonly written as: 6CO2 + 6H2O → C6H12O6 + 6O2. This means that the reactants, six carbon dioxide molecules and six water molecules, are converted by light energy captured by chlorophyll (implied by the arrow) into a sugar molecule and six oxygen molecules, the products.

Explanation:

3 0
3 years ago
THIS IS DUE TODAY!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASE HELP ME ASAPPPPPPPP CHEMISTRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
allochka39001 [22]

Answer:

\large \boxed{\text{0.1767 mol/L}}$

Explanation:

(a) Balanced equation

3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O  

(b) Moles of NaOH

\text{Moles of  NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}

(c) Moles of H₃PO₄

The molar ratio is 1 mol H₃PO₄:3 mol NaOH.

\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}

(d) Molar concentration of H₃PO₄

c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}

7 0
3 years ago
A 2.10g of unknown monoprotic acid is titrated with 38.10 mL of .265M NaOH. Calculate the molar mass of the ac? I already calcul
Shkiper50 [21]
V_{NaOH}=38,1mL=0,0381L\\
C_{m}=0,265M\\\\
n=C_{m}*V=0,265\frac{mol}{L}*0,0381L\approx0,0101mol

HR    +      NaOH ⇒ NaR + H₂O
1mol   :     1mol

0,0101mol \ \ \ \ \Rightarrow \ \ \  \ 2,1g\\
1mol  \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ m\\\\
m=\frac{1mol*2,1g}{0,0101mol}\approx 207,92g\\\\
M_{HR}=207,92\frac{g}{mol}
5 0
3 years ago
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