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4 years ago

A 2-kg bowling

2 answers:

Volgvan4 years ago

6 0

So for the first one It's 784.56 J (GPE)

For the second one it's KE is equal to 100 J. If you can give me a few more minutes I'll try to get the GPE

podryga [215]4 years ago

4 0

(05.01)

Which statement best explains if the graph correctly represents the proportional relationship y = 0.5x? (4 points)

<span>Yes, the points shown on the line would be part of y = 0.5xYes, all proportions can be shown on a graph of this lineNo, the points shown would not be part of y = 0.5xNo, proportions cannot be represented on a graph</span>

You might be interested in

A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 w

Stels [109]

Answer:

see explanation below

Explanation:

First to all, this is a redox reaction, and the reaction taking place is the following:

2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2

According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:

MhVh = MpVp

h would be the hydrogen peroxide, and p the permanganate.

But like it was stated before, the mole ratio is 5:2 so:

5MhVh = 2MpVp

Replacing moles:

5nh = 2MpVp

Now, we just have to replace the given data:

nh = 2MpVp/5

nh = 2 * 1.68 * 0.0213 / 5

nh = 0.0143 moles

Now to get the mass, we just need the molecular mass of the peroxide:

MM = 2*1 + 2*16 = 34 g/mol

Finally the mass:

m = 0.0143 * 34

m = 0.4862 g

8 0

3 years ago

A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)

Natalija [7]

Answer:

Explanation:

.98 is 98% of one and therefore they are missing 2%

4 0

3 years ago

Read 2 more answers

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

3 years ago

today, most of the world population growth is occurring in the world's poorest countries. Do you think carbon emissions will ten

laila [671]

Answer:Yes

Explanation:

3 0

3 years ago

Help fast !

jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0

3 years ago