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Law Incorporation [45]
3 years ago
13

What is the density of a 10 kg mass that occupies 5 liters? ( pls need help)

Chemistry
1 answer:
bulgar [2K]3 years ago
7 0

Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

10kg*\frac{1000g}{1kg} =10000 g

Now that we have grams, we can divide to get density.

d=\frac{10000g}{5 L}

d=2000g/L

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6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
deff fn [24]

Answer:

1120 gm

Explanation:

6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)

(a) Balance the equation.

(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete

combustion reaction?

FIRST, CORRECT THE EQUATION THEN BALANCE

2C2H6(G) + 7O2------------>  4CO2  + 6H2O

so for 10 moles of ethane, we need

7 X 5 = 35 MOLES O2

=35 MOLES O2

O2 HAS A MOLAR MASS OF 2X16 = 32 gm

35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm

6 0
3 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
3 years ago
How many molecules are there in 2.30g of NH3
Vinil7 [7]

Answer:

8.13x10^22 molecules

Explanation:

We can use the Avogadro's number(6.022 x 10^23 units / mole)

2.30 g NH3 (1 mol / 17.03 g ) (6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules

Hope this helps! Feel free to ask any questions!

6 0
2 years ago
Calculate the mass of 6 moles of H2O
RUDIKE [14]
<span>6mole of H2O contains 18×6 g=108 g</span>
8 0
3 years ago
Read 2 more answers
HELP: MODELING MOLECULES!!!!!!!!!!!!!!!!
garik1379 [7]
<span>Sulfur Hexachloride SCl6 So now we count the number of valence electrons each has by seeing what column it's in, (1-8) not counting the columns of the transition metals. Since Sulfur is in the 6th and Chlorine is in the 7th, and there are 6 chlorines, we can add up all their valence electrons: 6*1+7*6=48 valence electrons. But remember that electrons come in pairs, either in bonds or as lone pairs. So I usually divide the valence electron number by 2 and just think about placing pairs. It's up to you, but I think it's convenient since we can count "1" in our mind each time we place a bond or a electron pair. So we need to place 24 pairs/bonds. So we can guess that sulfur is a central atom and draw out a bond from sulfur to each chlorine. Since Sulfur is in the 3rd row it can use d-orbitals to break the octet rule. So when we bond all the chlorines onto sulfur we get:
(see the figure)

and 
</span><span>So we made 6 bonds, that means we used up 12 electrons, so if you're counting (AND YOU SHOULD BE!) you have 36 electrons or simply 18 electron pairs left to place. Now let's give chlorine a neutral charge.</span>

8 0
3 years ago
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