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alukav5142 [94]

2 years ago

8

I need help on this please

1 answer:

tino4ka555 [31]2 years ago

3 0

Answer:

The answer is c

Explanation:c

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Calculate the mass of CaCl2 ( formula wt. = 110.99) thatshould

NNADVOKAT [17]

Answer: 5622.6g

Explanation:

Note: Kf for water is 1.86°C/m.

The simple calculation is in the attachment below.

7 0

2 years ago

Read the two statements given below. Statement 1: All living matter at the smallest level is made of cells Statement 2: New anim

9966 [12]

Answer:

Statement 1: All living matter at the smallest level is made of cells

Explamation:

All living things are made of cells; the cell itself is the smallest fundamental unit of structure and function in living organisms.

Hope this helps

7 0

3 years ago

ANSWER FAST PLZ 25 POINTS!!!!!!!!!!!!!!!

Scilla [17]

Answer:

i think the answer is C)

Explanation:

8 0

3 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

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

$q_1=-q_2$

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

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

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

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

Determine the molarity of a solution with a volume of 435. mL and 0.550 mol of solute dissolved.

lora16 [44]

Answer:

M = 1.26

Explanation:

Molarity = mole of solution/liters of solution
435mL/1000 = .435L
Plugging in the numbers into the formula, we get:
Molarity = .550 mol/.435L = 1.26 M

8 0

1 year ago