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3 years ago

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Chemistry

1 answer:

Andrej [43]3 years ago

7 0

Answer:

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Explanation:

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Make a plot of arsenate species vs. pH for a 25 mM arsenate solution. Vary pH from 0-14. Neglect activity corrections. A groundw

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What compound are formed with lithium and selenium?

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Dates Modify 2020-11-15 Create 2005-08-08

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3 years ago

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What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

<u>Step 1 :</u> Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M

This means there is 0.05 mol per 100mL

<u>Step 2 </u>: calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

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3 years ago

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The word elliptical refers to what type of shape?

Ilya [14]

The word elliptical refers to an Oval.

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3 years ago

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4

Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = 560 J/(kg.K)

$c_1$ = specific heat of water = 4186 J/(kg.K)

$m_1$ = mass of iron = 825 g

$m_2$ = mass of water = 40 g

$T_f$ = final temperature of water and iron = ?

$T_1$ = initial temperature of iron = $352^oC=273+352=625K$

$T_2$ = initial temperature of water = $20^oC=273+20=293K$

Now put all the given values in the above formula, we get:

$(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)$

$T_f=537.12K$

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0

3 years ago