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3 years ago

Plz help! It takes a car going 40 mph 3 seconds to come to a stop. What is the acceleration?

Chemistry

1 answer:

lana66690 [7]3 years ago

3 0

Answer:

6 second???

Explanation:

Im so sorry if thats wrong!!!!

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What is the molarity of a NaCl solution containing 9.0<br> moles of NaCl in 3.0 L of solution?

ss7ja [257]

Answer:

3 M

Explanation:

Molarity equation: M = n/v

n = moles of solute

v = liters of solution

9 moles of NaCl / 3 L

9/3 = 3 M

4 0

3 years ago

Name the following compounds CaSe <br><br> don’t use common names such as water or salt

Contact [7]

Do it urself ;) jdhdhdhdhdhhdhdhdhdhdhdhdhdhgdhd

8 0

3 years ago

In the PhET simulation window, click the radio button labeled Mystery in the Blocks menu on the right-hand side of the screen. U

Yuliya22 [10]

Answer:

B, D, E, C, A

Explanation:

We have 5 blocks with their respective masses and volumes.

Block Mass Volume

A 65.14 kg 103.38 L

B 0.64 kg 100.64 L

C 4.08 kg 104.08 L

D 3.10 kg 103.10 L

E 3.53 kg 101.00 L

The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V

ρA = 65.14 kg / 103.38 L = 0.6301 kg/L

ρB = 0.64 kg / 100.64 L = 0.0064 kg/L

ρC = 4.08 kg / 104.08 L = 0.0392 kg/L

ρD = 3.10 kg / 103.10 L = 0.0301 kg/L

ρE = 3.53 kg / 101.00 L = 0.0350 kg/L

The order from least dense to most dense is B, D, E, C, A

4 0

2 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

When an atom is in the ground state what must happen for the atom to be in an excited state? what must happen for this atom to r

scoray [572]

Hi there

In order for an electron to jump into a higher energy state, it must first absorb energy (heat, light, etc).

When an electron goes back down to the ground state from the excited state, it emits energy usually in the form of a photon.

i hope this helps

4 0

3 years ago