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3 years ago

Katie throws a 3 kilogram ball fast at a speed of 15 meters what is the momentum of the ball katie threw

Chemistry

1 answer:

stiks02 [169]3 years ago

8 0

Answer:

Hey!

Your answer is 45kg.m/s

Explanation:

Using the formula P=M x V

(p=momentum...M=mass...V=velocity)

We do M x V

15 x 3

Which gives us an answer of 45kg.m/s!

HOPE THIS HELPS!!

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A student is planning a synthesis on a set of 3 alcohol isomers. A student wants to oxidize, then use the dried product of each

lyudmila [28]

Answer:

lollo

Explanation:

huh

5 0

2 years ago

In the combustion of propanol, what is the molar ratio of propanol (ch3ch2ch2oh) to co2? combustion of propanol: 2ch3ch2ch2oh +

maksim [4K]

Propanol molecular formula is C₃H₇OH
the balanced reaction for combustion of propanol is as follows;
2C₃H₇OH + 9O₂ --> 6CO₂ + 8H₂O
Molar ratio of reactants to products can be found out by the stoichiometry.
By looking at the coefficients of reactants and products in the balanced reaction equation, molar ratio can determined.
Therefore the molar ratio of propanol to CO₂ is 2:6, simplified --> 1:3

4 0

3 years ago

Which solution would most likely cause a plant placed in it to become firmer and more rigid? A. hypertonic B. hypotonic C. isoto

Anna11 [10]

"Hypotonic" is the one solution among the choices given in the question that would <span>most likely cause a plant placed in it to become firmer and more rigid. The correct option among all the options that are given in the question is the second option or option "B". I hope the answer has come to your great help.</span>

5 0

3 years ago

Read 2 more answers

As you have seen, ozone is formed in the upper atmosphere through the reaction:

baherus [9]

The rate law equation for Ozone reaction

r=k[O][O₂]

<h3>Further explanation</h3>

Given

Reaction of Ozone :.

O(g) + O2(g) → O3(g)

Required

the rate law equation

Solution

The rate law is a chemical equation that shows the relationship between reaction rate and the concentration / pressure of the reactants

For reaction

aA + bB ⇒ C + D

The rate law can be formulated:

$\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}$

where

r = reaction rate, M / s

k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹

a = reaction order to A

b = reaction order to B

[A] = [B] = concentration of substances

So for Ozone reaction, the rate law (first orde for both O and O₂) :

$\tt \boxed{\bold{r=k[O][O_2]}}$

5 0

3 years ago

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

NNADVOKAT [17]

First we need to calculate the number of moles of NaOH titrated.

molar concentration = number of moles / solution volume (liter)

number of moles = molar concentration × solution volume

number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH $_{3}$ -COOH + NaOH = CH $_{3}$ COONa + H $_{2}$ O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

mass = number of moles × molecular mass

mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100

concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %

3 0

2 years ago