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SVEN [57.7K]
3 years ago
7

Need help w the questions in the pic! thankyouu

Chemistry
1 answer:
katrin [286]3 years ago
8 0

Answer: Here are the answers

Explanation:

7a) Calcium Chloride +water+carbon dioxide

b)Calcium Oxide +Carbon dioxide

8a) zinc + carbon monoxide

b) Aluminum + Carbon dioxide

c) Lead + carbon dioxide

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Bryan and Alyssa are playing tug-of-war against Daniel and Maria. Bryan and Alyssa are exerting a combined force of 10 N. Daniel
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Answer:

D

Explanation:

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Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation y
sammy [17]

Answer:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Explanation:

We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

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Why does magma in the mantle rise through the crust above it
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BRAINLIESTTT ASAP!! PLEASE HELP ME :)
AlladinOne [14]

Answer : False, there will be two lithium and one oxygen atoms in a unit molecular structure of lithium oxide.

Explanation:

Electronic configuration of lithium is :

Li=1s^2s^1

In order to attain stable electronic configuration it will loose an electron and form positively charge cation.

Li^+=1s^22s^0

The electronic configuration of oxygen is:

O=1s^22s^22p^4

Oxygen being second most electronegative atom requires two electrons to attain noble gas configuration stability and form negatively charge ion with 2- charge.:

O^{2-}=1s^22s^22p^6

When two atom of lithium and oxygen comes together , one electron from each lithium atom get transferred to an oxygen atom which results in formation of lithium oxide.

2Li^++O^{2-}\rightarrow Li_2O

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3 years ago
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Determine the empirical formulas for compounds with the following percent compositions:
Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

3 0
3 years ago
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