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LenKa [72]
3 years ago
13

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

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Please help!! Which process occurs during all chemical reactions?
Sonbull [250]

Answer:

In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

Explanation:

please mark as brainliest if I helped you

3 0
3 years ago
What is the volume (L) occupied by a mole of an ideal gas,
irina [24]

Answer:

29.2 L

Explanation:

Given data

  • Moles (n): 1 mol
  • Pressure (P): 0.824 atm

626mmHg.\frac{1atm}{760 mmHg} =0.824atm

  • Temperature (T): 293 K

We can find the volume of the gas (V) of the using the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 1 mol × (0.0821 atm.L/mol.K) × 293 K / 0.824 atm

V = 29.2 L

7 0
3 years ago
Water can form a solution by mixing with:
MA_775_DIABLO [31]

Explanation:

Solid, liquids and gases

8 0
3 years ago
Read 2 more answers
1. A solution is made by mixing 50mL of 2.0M K2HPO4 and 25mL of 2.0M KH2PO4. The solution is diluted to a final volume of 200mL.
Kazeer [188]

Answer:

The pH of the final solution is 7.15

Explanation:

50 mL of 2.0 M of K_2HPO_4 and 25 mL of 2.0 M of KH_2PO_4 were mixed to make a solution

Final volume of the solution after dilution = 200 mL

Final concentration of K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M

Final concentration ofKH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M

We use Hasselbach- Henderson equation:

pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85

Substituting the values:

pH = 6.85+ log \frac{0.5}{0.25}pH = 6.85+ log 2pH = 6.85+ 0.3 = 7.15

Therfore,  the pH of the final solution is 7.15

5 0
3 years ago
PLEASE HELP :-)
timofeeve [1]
We can solve this problem using the long hand solution, wherein we 1 by 1 analyze the different equilibrium reactions or by simply using the Henderson Hasselbach equation. The equation is 

pH = -log(pKa) + log (salt/acid)

since the acid and the salt have the same concentration, the log (salt/acid) term is equal to zero. 

thus 

pH = -log(1.73*10-5)
pH = 4.76

please be careful with the negative sign 

8 0
4 years ago
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