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mart [117]
3 years ago
5

CO2 and H2O are both compounds. How do you know

Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

neutral groups of atoms formed by covalent bonds.

You might be interested in
Which substance has ionic bonds?<br><br><br><br>A) Br2<br><br>B) NH3<br><br>C) H2O<br><br>D) LiI
ludmilkaskok [199]
D

ionic bond is between a metal and non-metals

Lithium(Li) is a metal. it will lose one electron in ionization.
iodine (I)is non metal. it will gain one electron from Lithium in ionization

hope this helps
3 0
4 years ago
When the reaction shown is correctly balanced, the coefficients are: kclo3 → kcl + o2?
notka56 [123]
From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl 
coefficient 3 in front of oxygen molecule O2

We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
     KClO3 → KCl + O2
       K=1          K=1
       Cl=1         Cl=1
       O=3          O=2

We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen 
atoms on the right side to balance the oxygens:
     2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.

The number of potassium K atoms and chloride Cl atoms have changed as well:
     2KClO3 → KCl + 3O2
       K=2            K=1
       Cl=2          Cl=1
       O=6           O=6

We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side: 
     2KClO3 → 2KCl + 3O2, which is our final balanced equation.
        K=2           K=2
        Cl=2          Cl=2
        O=6           O=6
The potassium, chlorine, and oxygen atoms are now balanced.

5 0
3 years ago
Read 2 more answers
What is the total pressure of three gases, if the partial pressures are: 2.67 mmHg, 45.29 mmHg, and 789.6 mmHg
EastWind [94]

Answer:

The total pressure of three gases is 837.56 mmHg.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case, the total pressure can be calculated as:

PT= 2.67 mmHg + 45.69 mmHg + 789.6 mmHg

Solving:

PT= 837.56 mmHg

<em><u>The total pressure of three gases is 837.56 mmHg.</u></em>

7 0
3 years ago
One evening, Alex notes the Moon set in the west at around 8 PM. What would have been the approximate time of moonrise that day?
ra1l [238]

Answer:

C

Explanation:

Early evening about 5 p. m.

6 0
3 years ago
Read 2 more answers
a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


4 0
3 years ago
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