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3 years ago

7

la longitud de un cable de aluminio es de 30 m a 20 C. Sabiendo que el cable es calentado hasta 60 C y que el coeficiente de dil

atacion lineal del aluminio es de 24x 10 -51 C, determina la longitud final del cable y su dilatacion.

1 answer:

Grace [21]3 years ago

6 0

Answer:

30cm

Explanation:

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What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink

S_A_V [24]

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

<u>Answer:</u> 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

<u>Explanation:</u>

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

$0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol$

<u>Calculating volume of sweetened tea:</u>

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

$0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL$

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

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3 years ago

Aqueous solutions of ammonium oxalate and lithium acetate react.

Anarel [89]

Answer:

The reaction is a double displacement reaction

Explanation:

Let us consider the reaction equation of the reaction between ammonium oxalate and lithium acetate.

(NH4)2C2O4(aq) +2 CH3COOLi(aq) -------> 2NH4CH3CO2(aq) + Li2C2O4(s)

This is a displacement reaction. A double displacement reaction is a type of reaction in which two reactants exchange their ions to form two new compounds. Double displacement reactions usually lead to the formation of a solid product which is also called a precipitate.

The general form of a Double displacement reaction is of the format:

AB + CD → AD + CB

Where A,B,C and D represents different ions respectively.

A double displacement reaction can also be referred to as salt metathesis reaction, double replacement reaction, exchange reaction, or a double decomposition reaction, although the latter term is more strictly used when one or more of the reactants does not dissolve in the solvent.

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3 years ago

What is love and why are people scared to show love even if they like boys or girls

myrzilka [38]

Because society has shaped our mind on what we shall think about things

6 0

3 years ago

Read 2 more answers

2. When vinegar is added to eggshells, carbon dioxide gas is produced.Why?

skad [1K]

Answer:

same as his

Explanation:

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3 years ago

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

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3 years ago