Water comes from different a.levels b.phases c.sources d.stages

4. (08.01 MC)

Lelechka [254]

Answer:

It has a lower density in its solild stae than it does in its liquid state.

Explanation:

Ice floats, allowing life underneath to leave despite the top freezing.

6 0

3 years ago

A 0.500-g sample of chromium metal reacted with sulfur powder to give 0.963 g of product. Calculate the empirical formula of the

IgorC [24]

Answer: The empirical formula for the given compound is $Cr_2S_3$

Explanation : Given,

Mass of product = 0.963 g

Mass of Cr = 0.500 g

Mass of S = 0.963 g - 0.500 g = 0.463 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Cr = $\frac{\text{Given mass of Cr}}{\text{Molar mass of Cr}}=\frac{0.500g}{52g/mole}=0.00962moles$

Moles of S = $\frac{\text{Given mass of S}}{\text{Molar mass of S}}=\frac{0.463g}{32g/mole}=0.0145moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00962 moles.

For Cr = $\frac{0.00962}{0.00962}=1$

For S = $\frac{0.0145}{0.00962}=1.5$

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of Cr : S = 1 : 1.5

The ratio of Cr : S = 2 : 3

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cr : S = 2 : 3

Hence, the empirical formula for the given compound is $Cr_2S_3$

8 0

4 years ago

Which of the following element combinations will likely form polyatomic ions?

ArbitrLikvidat [17]

Oxygen and carbon are element combinations that will most likely form polyatomic ions.

6 0

4 years ago

What is Dalton's law of partial pressure

Leto [7]

If 3 substances be x,y,z

$\\ \tt\longmapsto P_{net}=P_x+P_y+P_z$

Where

$\\ \tt\longmapsto P_x=\dfrac{n_xRT}{V}$

$\\ \tt\longmapsto P_y=\dfrac{n_yRT}{V}$

$\\ \tt\longmapsto P_z=\dfrac{n_zRT}{V}$

P is partial pressure
n is number of moles
V is volume
T is temperature

7 0

3 years ago

Read 2 more answers

Calculate the molar solubility of CaCO3 in 0.250M Na2CO3 Kps CaCO3 is 4.96x10-9

Yuliya22 [10]

Answer:

solubility is 1.984x10⁻⁹M

Explanation:

When CaCO₃ is in water, the equilibrium that occurs is:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹

If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:

[Ca²⁺] [0.250M] = 4.96x10⁻⁹

Assuming you are adding an amount of CaCO₃:

[X] [0.250 + X] = 4.96x10⁻⁹

Where X is the amoun of CaCO₃ you can add, that means, solubility

X² + 0.250X - 4.96x10⁻⁹ = 0

Solving for X:

X = -0.25M → False answer, there is no negative concentrations.

X = 1.984x10⁻⁹M.

That means, solubility is 1.984x10⁻⁹M

3 0

3 years ago

Read 2 more answers

Water comes from different a.levels b.phases c.sources d.stages​

Water comes from different a.levels b.phases c.sources d.stages