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kobusy [5.1K]
3 years ago
11

Help, please X﹏X- The picture in the right is after the water drained out -

Chemistry
2 answers:
lutik1710 [3]3 years ago
6 0
There’s no picture in your question
ioda3 years ago
6 0

Answer:

a or c

Explanation:

Can I have the brainliest pls?

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A sample of water is heated from room temperature to just below the boiling point. The overall change in temperature is 72°C. E
Jlenok [28]

Answer : The correct option is, (a) 345 K

Explanation :

The conversion used for the temperature from degree Celsius to Kelvin is:

K=273.15+^oC

where,

K = temperature in Kelvin

^oC = temperature in centigrade

As we are given the temperature in degree Celsius is, 72

Now we have to determine the temperature in Kelvin.

K=273.15+^oC

K=273.15+(72^oC)

K=345.15\approx 345

Therefore, the temperature in Kelvin is, 345 K

7 0
4 years ago
Read 2 more answers
If 0.40 mol of H2 and 0.15 mol of O2 were to react as completely as possible to produce H2O what mass of reactant would remain?
liberstina [14]

Answer : The mass of reactant H_2 remain would be, 0.20 grams.

Solution : Given,

Moles of H_2 = 0.40 mol

Moles of O_2 = 0.15 mol

Molar mass of H_2 = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2H_2+O_2\rightarrow 2H_2O

From the balanced reaction we conclude that

As, 1 mole of O_2 react with 2 mole of H_2

So, 0.15 moles of O_2 react with 0.15\times 2=0.30 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

The moles of reactant H_2 remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant H_2 remain.

\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2

\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g

Therefore, the mass of reactant H_2 remain would be, 0.20 grams.

3 0
3 years ago
If you drop a ball and know how high it bounces what is Bounce height
blondinia [14]
36inches in the air first bounce second bounce 15
5 0
3 years ago
HELPPP!!! What is the volume of a 1.31 moles sample of gas if the pressure is 904 mmHg and the temperature is 37. degrees Celsiu
Tju [1.3M]

Answer:

27.99 dm³

Explanation:

Applying

PV = nRT................ Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

From the question, we were aksed to find V.

Therefore we make V the subject of the equation

V = nRT/P................ Equation 2

Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

V = (1.31×310×0.082)/(1.1897)

V = 27.99 dm³

4 0
3 years ago
Which compound can act as both a Bronsted-Lowry acid and a Bronsted-Lowry base?
Dvinal [7]
The answer is A. Water

Bronsted-Lowry base compounds are those that can accept protons

Bronsted-Lowry Acid Compounds are those that can recieve one

Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
5 0
3 years ago
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