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Alik [6]
3 years ago
10

A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 The temperature of the water increased fr

om 25 to 37°C. How much heat energy did the water absorb?
Chemistry
2 answers:
miss Akunina [59]3 years ago
6 0

Answer:

Find the solution's specific heat on a chart or use the specific heat of water, which is 4.186 joules per gram Celsius. Substitute the solution's mass (m), temperature change (delta T) and specific heat (c) into the equation Q = c x m x delta T, where Q is the heat absorbed by the solution.

Explanation:

olya-2409 [2.1K]3 years ago
3 0

Answer:

b

Explanation:

rdmjmnfh

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Answer:

But since the solubility product constant for each compound is provided, their relative solubility can be ranked from highest to lowest. Depending on the ranking above, it is evident that aluminum hydroxide Al(OH)3 A l ( O H ) 3 has the lowest solubility at 25 Celsius degreesAs temperature increases, its solubility increases as well. Notice, however, that it does not increase significantly. In fact, you can expect to be able to dissolve no more than 40 g of sodium chloride per 100 g of water at 80∘C

6 0
2 years ago
PLEASE HELP!!!
ladessa [460]

Answer:

the answer is c. [.4r]3d104324p

7 0
3 years ago
What does a chemical formula like Na2CO3 tells us?
USPshnik [31]

It has: 2 atoms of sodium (Na) 1 atom of Carbon 3 atoms of Oxygen bound together with ionic and polar-covalent bonds

8 0
4 years ago
How many liters are in 3.45 moles of oxygen gas at STP?
just olya [345]

Answer:

22.4L of one mole of any gas

or you can use PV=nRT

3.45*22.4=77.28

Explanation:

5 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
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