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MA_775_DIABLO [31]
3 years ago
10

Help Mepleaseeeee !!!

Chemistry
2 answers:
asambeis [7]3 years ago
7 0

Answer:

The 500kg boulder

Explanation:

Sindrei [870]3 years ago
7 0
500kg Boulder, the answer u have rn is right
You might be interested in
2Na+S→Na2S
Sedbober [7]

Answer:

2.0 moles S

Explanation:

To find the number of moles of S, you need to convert the moles Na to moles S via the mole-to-mole ratio. This ratio is represented by the coefficients in the balanced equation. Because you wish to find moles S, you want to put this number in the numerator. Because you want to eliminate the moles Na, this number should be in the denominator.

2 Na + 1 S ---> Na₂S

4.0 moles Na             1 mole S
----------------------  x  ---------------------  =  2.0 moles S
                                 2 moles Na

3 0
2 years ago
A sample of an unknown liquid has a volume of 30.0 mL and a mass of 6 g. What is its density? i need this hella quick
Lana71 [14]

density= mass/volume

density= 6g/30.0 ml

density= 0,2 g/ml

7 0
3 years ago
Which component in this mixture has the lowest density? A. glass. . B. water. . C. sawdust. . D. sand
Yuri [45]
The component in the mixture with the lowest density would be the component floating on the top of the mixture, i.e. the lightest in weight. Therefore, it would be the sawdust.
3 0
3 years ago
Read 2 more answers
Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th
Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>

6 0
3 years ago
A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system
AleksAgata [21]
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w

Clearing the specific heat of the unknown substance:

c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}

c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}
6 0
3 years ago
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