Find the percent composition of water in barium chloride dihydrate. Round your answer to the

QUESTION 1 (POLL)

Airida [17]

Answer:

They stay the same

Explanation:

4 0

2 years ago

In the Formula 3H2O what is the 3 called

vekshin1

Its very simple .

3H2O ---- three H two O

Its 3 molecules of water

3 0

2 years ago

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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o

yaroslaw [1]

<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

Mass of the unknown metal sample as 58.932 g
Initial temperature of the metal sample as 101°C
Final temperature of metal is 23.68 °C
Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g
Initial temperature of water = 21°C
Final temperature of water is 23.68 °C
Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

= 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q = m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

= 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>

We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.

Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0

2 years ago

2Mg + O2 → 2MgO<br><br> If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

LekaFEV [45]

<h3>Answer:</h3>

9.6724 g MgO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})$
Multiply/Divide: $\displaystyle 9.67241 \ g \ MgO$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

9.67241 g MgO ≈ 9.6724 g MgO

5 0

2 years ago

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Write the balanced chemical equation for the following: The destruction of a marble statuary by acid rain: aqueous nitric acid r

Mashcka [7]

Answer: The coefficient in front of nitric acid is 2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas. Liquids are represented by (l) and gases are represented by (g).

The balanced chemical reaction is :

$CaCO_3(s)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+CO_2(g)+H_2O(l)$

7 0

2 years ago