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Pavlova-9 [17]
2 years ago
12

A hockey stick is in contact with a 165-g puck for 22.4 ms; during this time, the force on the puck is given approximately by f(

t)=a+bt+ct2, where a =

Chemistry
1 answer:
mario62 [17]2 years ago
3 0
I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:

F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²

a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.

F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>

b.) To solve for the distance, the formula is:

d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>

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Write a balanced half-reaction describing the oxidation of gaseous dihydrogen to aqueous hydrogen cations.
xz_007 [3.2K]
The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
8 0
2 years ago
Read 2 more answers
What is the empirical formula of a compound which contains 84.4% c and 15.6% h by mass?
podryga [215]

Answer: The empirical formula for the given compound is CH_2

Explanation : Given,

Percentage of C = 84.4 %

Percentage of H = 15.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 84.4 g

Mass of H = 15.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{84.4g}{12g/mole}=7.03moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{15.6g}{1g/mole}=15.6moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.

For Carbon = \frac{7.03}{7.03}=1

For Hydrogen  = \frac{15.6}{7.03}=2.22\approx 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given compound is C_1H_2=CH_2

7 0
2 years ago
The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH . It requires 11
Solnce55 [7]

Answer:

0.190 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH = NaCl + H2O

11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:

0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol

The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.

1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:

M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M

5 0
2 years ago
Read 2 more answers
An increase in temperature will effect vapor pressure by:
-BARSIC- [3]

Answer: Increases.

Explanation:  As the temperature of a liquid or solid increases its vapor pressure also increases. Conversely, vapor pressure decreases as the temperature decreases.

5 0
2 years ago
Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
2 years ago
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