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butalik [34]
3 years ago
12

Hi how are u guys :)

Chemistry
2 answers:
xxTIMURxx [149]3 years ago
7 0

Answer:

good just bored and you :)

gtnhenbr [62]3 years ago
3 0

Answer:good

Explanation:

You might be interested in
Which of these requires accurate coefficients in a reaction?
aksik [14]

Answer:

A: molar ratio

Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.

6 0
3 years ago
Find the formula for the hydrate<br>0.737 g MgSO3 and 0.763 g H2O
babunello [35]

The required formula of hydrate is MgSO₃.6H₂O.

<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
n = W/M, where

  • W = given mass
  • M = molar mass

Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol

Moles of H₂O = 0.763g / 18g/mol = 0.04 mol

Number of H₂O molecule = 0.04/0.007 = 5.7 = 6

So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

To know more about hydrate compound, visit the below link:

brainly.com/question/22411417

#SPJ1

6 0
2 years ago
Does the tempurture of the vinegar affect the volcano flow
aleksandr82 [10.1K]
The temperature of vinegar does increase the rate of reaction according to the collision theory.

Hope this answers your question!


6 0
2 years ago
What is the mass of substance that exactly fills a 247.2 mL container? The density of the substance is 0.81 g/mL. Give the answe
matrenka [14]

Answer:

<h2>mass = 200.23 g</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

Since we are finding the mass

<h3>mass = Density × volume</h3>

From the question

Density = 0.81 g/mL

volume = 247.2 mL

Substitute the values into the above formula and solve for the mass

mass = 0.81 × 247.2

= 200.232

We have the final answer as

<h3>mass = 200.23 g to 2 decimal places</h3>

Hope this helps you

3 0
3 years ago
Question 2 of 50
wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0
3 years ago
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