All categories

2 years ago

What is the molarity of a solution that is prepared by adding 3 moles of CaSO4 in 1.5 L of water?

Chemistry

1 answer:

blondinia [14]2 years ago

6 0

Answer:

$[CaSO_4] = 2 mol/L$

Explanation:

$[CaSO_4] = \frac{3 mol}{1.5 L} = 2 mol/L$

You might be interested in

The balanced equation below shows the products that are formed when butane (C4H10) is combusted.

Nataly [62]

Answer:

2:8

Explanation:

The reaction equation is a given as:

2C₄H₁₀ + 130₂ → 8CO₂ + 10H₂O

From the reaction equation, the mole ratio is 2:8

Butane is C₄H₁₀

Carbon dioxide CO₂

From the reaction;

2 moles of butane will produce 8 moles of carbon dioxide

3 0

3 years ago

Read 2 more answers

A 0.8kg object displaces 500ml of water what is its specific gravity

worty [1.4K]

Specific gravity is the ratio of density of substance and density of water

We know that density of water = 1 g /mL at standard conditions

now as given that the 0.8 Kg of the substance / object is able to displace 500mL of water , it means that

Mass of object = 800g

The volume occupied by 800g of object = 500 mL

Density = mass / volume

Density of object = 800 / 500= 1.6 g / mL

The specific gravity of object = density of object / density of water = 1.6 / 1 = 1.6 (no units)

3 0

3 years ago

What would the products be for the reaction between Na3PO4 + MgSO4?

ivann1987 [24]

MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2

Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2

Explanation:

7 0

3 years ago

Magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide as shown.

lorasvet [3.4K]

C. 21.0g

that’s the aweeee

7 0

3 years ago

Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate st

Ipatiy [6.2K]

Answer:

(3R,4R)-4-bromohexan-3-ol

Explanation:

In this case, we have reaction called halohydrin formation. This is a markovnikov reaction with anti configuration. Therefore the halogen in this case "Br" and the "OH" must have different configurations. Additionally, in this molecule both carbons have the same substitution, so the "OH" can go in any carbon.

Finally, in the product we will have chiral carbons, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)

I hope it helps!

5 0

3 years ago