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wel
3 years ago
14

Hi......................... HELLO

Chemistry
2 answers:
GREYUIT [131]3 years ago
7 0

Answer:

Hii hello how are you....

PtichkaEL [24]3 years ago
6 0

Answer:

hi

Explanation:

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2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?
sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

Q = c \times m \times \Delta T

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

m = \frac{Q}{c \times \Delta T  }  = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C)  } = 672 g

The mass of water that can be warmed under these conditions is 672 grams.

5 0
3 years ago
What kind of chemical change do you think occurs when a banana peel turns brown in the open air?
sineoko [7]
It has to be oxydation. Oxygen is what slowly poisons us until we die and makes us wrinkle.
4 0
4 years ago
Read 2 more answers
Which reactant will be in excess, and how many moles of it will be left over?
timofeeve [1]

Answer:

Oxygen with 0.36 moles left over

Explanation:

5 0
3 years ago
Read 2 more answers
The diagram below shows how heat can be transferred in the atmosphere as hot air rises and cool air sinks. When the hot, wet air
ch4aika [34]

Answer:

d) convection

Explanation:

7 0
3 years ago
A buffer is prepared by adding 150mL of 0.50 M NH3 to 250mL of 0.50 M NH4NO3. What is the pH of the final solution? (Kb for NH3
Juli2301 [7.4K]

From the calculations, the pH of the final solution is 9.04.

<h3>What is the pH of the buffer?</h3>

We can use the Henderson Hasselbach equation to obtain the final pH of the solution in terms of the pKb and the base concentration.

Number of moles of salt = 250/1000 L * 0.5 M = 0.125 moles

Number of moles of base = 150/1000 L * 0.5 M = 0.075 moles

Total volume of solution = 250ml + 150ml = 400ml or 0.4 L

Molarity of base = 0.075 moles/ 0.4 L = 0.1875 M

Molarity of salt = 0.125 moles/ 0.4 L = 0.3125 M

pOH = pKb + log[salt/base]

pKb = -log(1.8 x 10^-5) = 4.74

pOH = 4.74 + log[0.3125/0.1875 ]

pOH = 4.96

pH = 14- 4.96

pH = 9.04

Learn more about pH:brainly.com/question/15289741

#SPJ1

5 0
2 years ago
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