During a process called photoact, ________ give up an electron as a part of the light-dependent reactions.

Using only the following elements P, Br, and Mg, give the formulas for:A. an ionic compound. B. a molecular compound with polar

Talja [164]

Answer:

<h2>1. Ionic compound- $MgBr_2$ </h2><h2>2. Polar molecular compound- $PBr_3$ </h2>

Explanation:

Mg is a metal that has 12 atomic numbers and thus its electronic configuration is $1s^22s^22p^63s^2$ . The outer most shell of this element has 2 electrons so it loses 2 electrons and thus form $Mg^2^+$ ions. Br is a nonmetal and has 35 atomic number so its electronic configuration is $1s^22s^22p^63s^23p^64s^23d^1^04p^5$ . Since its outermost shell has 7 electrons so it can accept one electron and thus forms $Br^-$ . So magnesium ion and bromide ion combine and forms an ionic compound $MgBr_2$ .

P is also a nonmetal and combine with Br with covalent bond and due to electronegativity differences form polar covalent compound such as $PBr_3$ .

8 0

2 years ago

Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

Activity Two: Just Lemons, Inc. Production

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.

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

Water 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

Just Lemons Lemonade Recipe Equation:

2 water + sugar + lemon juice = 4 lemonade

Mole conversion factors:

1 mole of water = 1 cup = 236.59 g

1 mole of sugar = 1 cup = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

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

Water:

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.

Sugar:

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.

Lemon juice:

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

How many liters of 3.0 M NaOH solution will react with 2.4 mol H2SO4? (Remember to balance the equation.)

timurjin [86]

Answer:

1.6 L is the volume of NaOH that has reacted

Explanation:

The balanced reaction is:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

This is a neutralization reaction between a strong acid and a strong base. The products are the correspond salt and water.

We propose this rule of three:

1 mol of sulfuric acid needs 2 mol of NaOH to react to react

Then, 2.4 moles of H₂SO₄ will react with (2.4 . 2) / 1 = 4.8 moles of NaOH

As molarity is 3M, we can determine the volume of our solution

Molarity (M) = mol / volume(L) → Volume(L) = mol / Molarity

Volume(L) = 4.8 mol / 3 M = 1.6 L

3 0

2 years ago

A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.

prohojiy [21]

The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7

7 0

2 years ago

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?

Xelga [282]

let the actual height of car be x

now, according to question,