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1 year ago

What happens to carbon dioxide?

Chemistry

2 answers:

oksian1 [2.3K]1 year ago

5 0

Carbon moves from the atmosphere to the oceans.

Molodets [167]1 year ago

3 0

Moves from the atmosphere to the oceans

You might be interested in

Where do you place the symbols that represent the state of matter in an equation?

nexus9112 [7]

Subscript in a pair of brackets

5 0

2 years ago

Which total mass of iron is necessary to produce 1.00 mole of copper?

Harman [31]

1 mole Fe produces 1 mole Cu
<span> molar mass of Fe is 55.8 g / mole

answer: </span><span>55.8g</span>

4 0

3 years ago

What are the two types of collisions? What is the difference between them? Types: a. Types: b. Compare and contrast the two type

mr_godi [17]

Answer:

The two types of collisions are :

Type a)

<u>Elastic collision</u>

Type b)

<u>Inelastic collision</u>

Explanation:

Collision : It is the event when two bodies collide with each other for small period of time.

During collision , the bodies exert force to each other.

Example :

When boxer hits with punches .

When bat hits the ball in cricket match.

So, collision is short duration interaction of two objects. When the objects collides , there is change in their velocity.

All collision follow law of conservation of momentum . Their type is decided by , whether they follow conservation of energy also.

<u>Compare and contrast the two types</u>

a) Elastic collision : Those collision in which no loss or gain of kinetic energy will occur. They follow conservation of kinetic energy. Example : ideal gaseous molecule

b) Inelastic collision : Those collision in which Change in kinetic energy will occur. They do not follow conservation of kinetic energy.Almost all conservation are inelastic.

Here Kinetic energy get converted into other form of energy.

5 0

2 years ago

How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol

SVEN [57.7K]

<h3>Answer:</h3>

0.0157 g Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u /> $3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} )$ = 0.015733 g Au

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.015733 g Au ≈ 0.0157 g Au

3 0

2 years ago

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$

Therefore, the leftover of sodium nitrate is:

$m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3$

Finally, the percent yield is computed via:

$Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%$

Best regards!

6 0

2 years ago