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vredina [299]
4 years ago
9

when you increased the ramp height did the stopping distance change as compare with the lower ramp height

Chemistry
1 answer:
ExtremeBDS [4]4 years ago
8 0
If a ball is running down a ramp, why is it that when you change the height of the ramp, the ... of the ramp, then you will increase the acceleration of a ball which rolls down the ramp.
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If a gas sample occupies 455 mL at 25°C and 744 torr, what will its volume be at STP?
ozzi

Answer: It can be written as: V = nRT/P. "P" is pressure, "V" is volume, n is the number of moles of a gas, "R" is the molar gas constant and "T" is temperature. Record the molar gas constant "R". R = 8.314472 J/mole x K.

Explanation: That the answer

7 0
3 years ago
3- a) write the formula unit of calcium fluoride.
zvonat [6]

Answer:

a) CaF₂.

b) 7.81g of CaF₂ are present.

Explanation:

a) The calcium ion has as charge Ca²⁺ and fluoride ion is F⁻, that means formula unit is:

CaF₂

b) 1 molecule of CaF₂ contains 2 anions, F⁻. Thus, the moles of CaF₂ is:

1.2x10²³ anions * (1 moleculeCaF₂ / 2 anions) = 6x10²² molecules of CaF₂

6.022x10²³ molecules = 1mol:

6x10²² molecules of CaF₂ * (1mol / 6.022x10²³molecules) = 0.10 moles CaF₂.

1 mole of CaF₂ has a mass of 78.07g:

0.10 moles CaF₂ * (78.07g / mol) =

7.81g of CaF₂ are present

5 0
3 years ago
86.64 g LiBr mol LiBr
Otrada [13]

Explanation:

I don't think your question is correct, please post the right question in the comment box.

6 0
3 years ago
Read 2 more answers
At 66.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
andreev551 [17]

Answer:

\boxed{5.5 \times 10^{-28}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \frac{RT}{zF}\ln Q

Step 1. Calculate E°

SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ ⇌ SO₂(g) + 2H₂O(ℓ)

<u>2Br⁻⇌ Br₂(aq) + 2e⁻                                                                </u>

SO₄²⁻(aq) + 4H⁺(aq) + 2Br⁻(aq) ⇌ Br₂(aq) + SO₂(g) + 2H₂O(ℓ)

E° = 0.17 - 1.0873 = -0.92 V

Step 2. Calculate Q

E   =  0      V

E° = -0.92 V

R = 8.314 J·K⁻¹mol⁻¹

T = 66 °C

n = 2

F = 96 485 C/mol

Calculations:  

T = 66.0 + 273.15 = 339.15 K

0 = -0.92 - \dfrac{8.314 \times 339.15}{2 \times 96 485}\ln Q\\\\0.92 = -0.0146\times \ln Q \\\\\ln Q = -\dfrac{0.92}{0.0146} = -62.8\\\\Q = e^{-62.8} =\boxed{5.5 \times 10^{-28}}\\

6 0
4 years ago
Select the two chemical reactions that adhere to the law of conservation of matter.
rodikova [14]

####3 ##################### 2AS###################

6 0
2 years ago
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