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garik1379 [7]
3 years ago
6

Balance this equation: Ca(OH)2 + HCI → CaCl2 + H20

Chemistry
1 answer:
Verizon [17]3 years ago
5 0

Answer:

Double Displacement (Acid-Base)

Explanation:

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1. A sample of gas has an initial volume of 25 L and an initial pressure of 123.5 kPa. If the pressure changes to
sveta [45]

Answer:

\large \boxed{\text{17.mL}}

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{123.5 kPa}\qquad & V_{1} &= & \text{25 L} \\p_{2}& =& \text{179.9 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:  

\begin{array}{rcl}123.5 \times 25 & =& 179.9V_{2}\\3088 & = & 179.9V_{2}\\V_{2} & = &\dfrac{3088}{179.9}\\\\& = &\textbf{17 L}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{17 L}}

6 0
3 years ago
Answers for cardiovascular system crossword puzzle
KengaRu [80]
I dont know the answer

5 0
3 years ago
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A 5.0-gram sample of Fe(s) is to be placed in 100. milliliters of HCl(aq). Which changes will result in the fastest rate of reac
sveta [45]
The answer is <span>increasing the surface area of Fe(s) and increasing the concentration of HCl(aq). 

the rate of reaction can be increased if the interaction between the reactants in better. by increasing the surface area of a solid, such as Fe, you make more possible each molecule of Fe to interact with HCl. Also, by increasing the concentration of HCl, it mean there is more molecule of HCl to interact with Fe.</span>
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3 years ago
If there were 10 grams of a radioactive isotope, how much of the sample would be left after 1 half-life? How much after 2 half-l
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One half-life: 5 grams
Two half-lives: 2.5 grams
3 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
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