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1 year ago

13

Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50μm2. Convert this

to square meters

1 answer:

Salsk061 [2.6K]1 year ago

8 0

You determine that a certain layer of graphene covers an area of 1.50μm2. ... Convert this to square meters. 1.

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How many moles of CaCO3 are needed to get 2.35 moles of CaO

GrogVix [38]

I'm pretty sure its 30.6 grams of CA

7 0

2 years ago

A hockey stick is in contact with a 165-g puck for 22.4 ms; during this time, the force on the puck is given approximately by f(

mario62 [17]

I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:

F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²

a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.

F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>

b.) To solve for the distance, the formula is:

d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>

3 0

1 year ago

Ammonia has a Kb of 1.8 × 10−5. Find [H3O+], [OH−], pH, and pOH for a 0.310 M ammonia solution.

lawyer [7]

Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

pH = 11.4; pOH = 2.6

Explanation:

The chemical equation is

$\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$$

For simplicity, let's re-write this as

$\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$$

1. Calculate [OH]⁻

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

0.310 0 0

-x +x +x

0.310-x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}$

2. Calculate the pOH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}$

3. Calculate the pH

$\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}$

4 Calculate [H₃O⁺]

$\text{H$_{3}$O$^{+}$} = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}$

5 0

1 year ago

Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

Carbon dioxide, CO₂, changes from the solid phase to the gas phase at 1 atm and 194.5 K. In the solid phase, CO₂ is often called

FinnZ [79.3K]

Answer:

1. From water vapor to the dry ice;

2. The potential energy is higher before the water vapor condenses;

3. The thermal energy is higher in the 2.0 kg block.

Explanation:

1. The heat flows from the system with high temperature to the system with low temperature. The water vapor is at 298 K, and the dry ice is at 194.5 K.

2. The energy of the molecules is related to the temperature and the physics state. At the gas state, the molecules are more agitated, and the energy is higher than the liquid state. So, when the vapor condenses to a liquid, the energy decreases.

3. The thermal energy can be calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat, and ΔT the variation in the temperature. So, when the mass increase, thermal energy also increases.

5 0

1 year ago