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2 years ago

Which one of these 3 ( beta, alpha, gamma) is the strongest and which is the weakest?

Chemistry

1 answer:

prisoha [69]2 years ago

8 0

Answer: beta, gamma, alpha

Explanation: Beta is weakest gamma is middle alpha is strongest

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Sodium metal is sometimes used as a cooling agent in heat exchange units because of its relatively high molar heat capacity of 2

IgorC [24]

Answer:

The specific heat of sodium is 1,23J/g°C

Explanation:

Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.

starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C

$28,2 \frac{J}{mol C} x \frac{1mol}{23g} = 1,23 J/g C$

4 0

2 years ago

What would be the new volume of a gas if 455 of that gas is cooled from 30.0°C to -80.0°C?

Lapatulllka [165]

Answer:

44555

Explanation:

is the answer

5 0

2 years ago

Read 2 more answers

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

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

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity 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.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

7 0

2 years ago

An acid sample of an unknown concentration is contained in a erlenmeyer flask. Which technique below would be best suited to ana

Ivenika [448]

Answer:

Titration

Explanation:

The best technique which can be used to determine the number of moles of the HCl in the sample is titration.

The given amount of HCl solution must be titrated with known concentration of the base like NaOH.

The volume of NaOH required must be noted also.

According to the reaction,

$NaOH+HCl\rightarrow NaCl+H_2O$

At equivalence point

Moles of $HCl$ = Moles of $NaOH$

Considering:-

Moles of $HCl=Molarity_{NaOH}\times Volume_{NaOH}$

Thus, in this way, moles of HCl can be determined.

4 0

2 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

4 0

2 years ago