The highest tides occur when:

Which of the following is an example of acceleration?

anzhelika [568]

Answer:

it's either B. or C.. hope this helps!

Explanation:

4 0

3 years ago

Read 2 more answers

I need answers to question 1,2,3

sashaice [31]

Answer:

1. 0.125 mole

2. 42.5 g

3. 0.61 mole

Explanation:

1. Determination of the number of mole of NaOH.

Mass of NaOH = 5 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mole of NaOH =?

Mole = mass /molar mass

Mole of NaOH = 5/40

Mole NaOH = 0.125 mole

2. Determination of the mass of NH₃.

Mole of NH₃ = 2.5 moles

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mass of NH₃ =?

Mass = mole × molar mass

Mass of NH₃ = 2.5 × 17

Mass of NH₃ = 42.5 g

3. Determination of the number of mole of Ca(NO₃)₂.

Mass of Ca(NO₃)₂ = 100 g

Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]

= 40 + 2[14 + 48]

= 40 + 2[62]

= 40 + 124

= 164 g/mol

Mole of Ca(NO₃)₂ =?

Mole = mass /molar mass

Mole of Ca(NO₃)₂ = 100 / 164

Mole of Ca(NO₃)₂ = 0.61 mole

6 0

3 years ago

Why are most household items made up of oxygen?

Alchen [17]

It is in our atmosphere

5 0

3 years ago

A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t

kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0

3 years ago

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol

defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $\frac{[A^{-}]}{[HA]}$

1,38 = $\frac{[A^{-}]}{[HA]}$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $\frac{60,05 g}{1mol}$ = 9,0 g of acetic acid

I hope it helps!

5 0

3 years ago