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Alex
3 years ago
8

What is the answer to this question What is H2O

Chemistry
2 answers:
BigorU [14]3 years ago
8 0
H2O is a water molecule.
Inessa05 [86]3 years ago
8 0
It's a water molecule
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A nucleus of mass number 81 contains 46 neutrons. an atomic ion of this element has 36 electrons in it. write the symbol for thi
nasty-shy [4]

A nucleus of mass number 81 contains 46 neutrons

Mass number=number of protons+number of neutrons

81=number of protons+ 46

number of protons=81-46

number of protons=35

In an atom, number of electrons=number of protons=35

So, the element is ⁸¹Br₃₅ .

Atomic ion of Br, having 36 electrons in it will be:

                                                                                      ⁸¹Br⁻₃₆

The symbol for this atomic ion will be Br and charge on the ion will be -ve.



8 0
3 years ago
A home solar energy unit stores energy for later use. On a sunny day, the temperature of the water in the tank increased from 23
igomit [66]

Answer:

391.46 L

Explanation:

Given:

Initial temperature of the water = 23.4°C

Final temperature of the water = 39°C

Temperature change for the water, ΔT = (39 - 23.4) = 15.6°C

Heat absorbed by the water = 2.55 × 10⁴ kJ = 2.55 × 10⁷ J

Density of the water = 0.998 g/mL

Specific heat of the water, C = 4.184 J/g°C

Now,

The heat absorbed by the water, q = mCΔT

where, m is the mass of the water

Thus,

we have

2.55 × 10⁷ J = m × 4.184 × 15.6

or

m = 390682.45 grams

also,

Density = mass / volume

thus,

0.998 = 390682.45 / volume

or

Volume = 391465.38 mL

or

Volume = 391.46 L

7 0
4 years ago
What is the moment for the left hand side of this balance?
aleksklad [387]

Answer:

18

Explanation:

6 (weights) x 3 (distance) = moment

5 0
3 years ago
#1: At STP, how many molecules of nitrogen gas are in 22.4 L?
OLEGan [10]

At STP, 1 mol = 22.4

1 mol = 6.022 X 10^ 23

so your answer is right it is B 6.022 X 10^ 23

3 0
3 years ago
A hamburger containing 335.4 kcal of energy was combusted in a bomb calorimeter with an unknown heat capacity. The temperature o
Zielflug [23.3K]

Answer:

Cv_{calorimeter}=18.4kcal/K

Explanation:

Hello!

In this case, since the combustion of the hamburger released 335.4 kcal of energy and that energy is received by the calorimeter, we can write:

Q_{hamburguer}=-Q_{calorimeter}

And the heat of the calorimeter is written in terms of the temperature change and the calorimeter constant:

Q_{hamburguer}=-Cv_{calorimeter}\Delta T

Thus, given the released heat by the hamburger due to its combustion and the temperature change, Cv for the calorimeter turns out:

Cv_{calorimeter}=\frac{-Q_{hamburguer}}{\Delta T} =\frac{-(-335.4kcal)}{18.2K}\\\\Cv_{calorimeter}=18.4kcal/K

Best regards!

4 0
3 years ago
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