Could someone answer this

I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP

GalinKa [24]

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

3 0

2 years ago

Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete

oee [108]

Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.

5 0

2 years ago

Please select the word from the list that best fits the definition

sammy [17]

Pretty sure it’s ,B. Microscope!

8 0

2 years ago

How can a scientist make sure that the results are not a mistake in an experiment?

olga55 [171]

Repeat trials multiple times

8 0

3 years ago

2AlF3 + 3K2O → 6KF + Al2O3<br><br> How many grams of AlF3would it take to make 15.524 g of KF?

mr_godi [17]

<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0

2 years ago

Read 2 more answers