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2 years ago

30)

Chemistry

1 answer:

Feliz [49]2 years ago

6 0

Answer:

D has to be based on facts so d would be the answer and idea that can only be proven true those are facts not hypothesis or guesses

Explanation:

based on a body of facts that have been repeatedly confirmed through observation and experiment. Such fact-supported theories are not "guesses" but reliable accounts of the real world."

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Explain how intermolecular attractions between molecules influence the bulk of properties of a material

irina1246 [14]

Answer:

Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid.

5 0

2 years ago

Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how

Afina-wow [57]

Answer:

11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

Explanation:

1. The balanced chemical equation is the following:

$Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)$

2. Use the molar mass of the $H_{2}O$ , the molar mass of the $NH_{3}$ and the stoichiometry of the balanced chemical reaction to find how many grams of $NH_{3}$ are produced:

Molar mass $H_{2}O$ = 18 $\frac{g}{mol}$

Molar mass $NH_{3}$ = 17 $\frac{g}{mol}$

$36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}$

Therefore 11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

3 0

2 years ago

KOH(Ac) + HNO3 → KNO3(Ac) + H2O cual es su reacio

Bumek [7]

Answer:

corbonization for dictionnal

Explanation:

correct

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2 years ago

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In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so

gulaghasi [49]

Answer: The moles of hydroxide ions present in the sample is 0.0008 moles

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0

2 years ago