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Aleksandr-060686 [28]
2 years ago
8

PLEASE HELP MEEEEEEEEEEEE

Chemistry
1 answer:
dlinn [17]2 years ago
5 0

Answer:

DNA.....................I hope

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The specific heat of ethanol is 2.44 J/g ֯C. How many kJ of energy are required to heat 50.0 grams of ethanol from -20 ֯C to 68
Aliun [14]

Answer:

Heat energy required (Q) = 10.736 KJ

Explanation:

Given:

Specific heat of ethanol (C) = 2.44 J/g °C

Mass of ethanol (M) = 50 gram

Initial temperature (T1) = -20°C

Final temperature (T1) = 68°C

Find:

Heat energy required (Q) = ?

Computation:

Change in temperature (ΔT) = 68°C - (-20°C)

Change in temperature (ΔT) = 88°C

Heat energy required (Q) = mC(ΔT)

Heat energy required (Q) = (50)(2.44)(88)

Heat energy required (Q) = 10,736 J

Heat energy required (Q) = 10.736 KJ

5 0
3 years ago
What volume of 12 M HCl solution is needed to make 2.5 L of 1.0 M HCI?
Jlenok [28]

Answer:

0.21 L of 12 M HCL

Explanation:

CV=CV

(12)(x)=(1.0)(2.5)

x=0.21

8 0
3 years ago
1. At room temperature, air is usually a (solid, liquid, gas).​
kirza4 [7]
At room temperature the air is Gas
3 0
2 years ago
Ik i already asked this but i need diff point of views
salantis [7]

Answer:

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5 0
2 years ago
Read 2 more answers
calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg
alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

learn more about Ions:

brainly.com/question/13692734

#SPJ4

6 0
1 year ago
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