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3 years ago

Please answer 11 & 12 and explain why ..

Chemistry

1 answer:

irga5000 [103]3 years ago

3 0

11.It is reffered to as the blanket around the atmosphere because it raps in all the air
12.water can never dissapear as long as we keep digesting and drinking water

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CAN SOMEONE HELP PLEASE!!!!

miv72 [106K]

Answer: The Hubble

Explanation:

8 0

3 years ago

Read 2 more answers

How many milliliters of 2.5 M HCl are required to exactly neutralize 1.5 L of 5.0 M NaOH?

Airida [17]

Answer:

3000mL

Explanation:

The following data were obtained from the question:

Volume of acid (Va) =..?

Molarity of acid (Ma) = 2.5M

Volume of base (Vb) = 1.5L

Molarity of base (Mb) = 5M

Next, we shall write the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

We obtained the following:

Mole ratio of the acid (nA) = 1

Mole ratio of base (nB) = 1

Next, we shall determine the volume of HCl needed for the reaction. This can be obtained as follow illustrated below:

MaVa / MbVb = nA/nB

2.5 x Va / 5 x 1.5 = 1

Cross multiply

2.5 x Va = 5 x 1.5

Divide both side by 2.5

Va = 5 x 1.5 / 2.5

Va = 3L

The volume of the acid required is 3L.

Finally, we shall convert the volume of the acid from litre (L) to millilitre (mL). This is illustrated below:

1L = 1000mL

Therefore, 3L = 3 x 1000 = 3000mL.

Therefore, the volume of the acid, HCl in mL needed fi4 the reaction is 3000mL

7 0

4 years ago

* The measurement for volume is? O grams O liters Oatom O mole

lord [1]

Answer: cubic units

6 0

3 years ago

What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 moles of

IRINA_888 [86]

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of $V_2O_5$ = 1.0 moles

Moles of $Ca$ = 4.0 moles

According to the given reaction:-

$V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}$

1 mole of $V_2O_5$ react with 5 moles of $Ca$

Moles of Ca available = 4.0 moles

Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)

The formation of the product is governed by the limiting reagent. So,

5 moles of Ca on reaction forms 2 moles of V

1 mole of Ca on reaction for 2/5 mole of V

4.0 mole of Ca on reaction for $\frac{2}{5}\times 4$ mole of V

Moles of V = 1.6 moles

Theoretical yield of vanadium = 1.6 moles

4 0

3 years ago

If you have 3 moles of FeCl3, how many moles of MgCl2 can be produced? FeCl3 + Mg --> MgCl2 + Fe

oksian1 [2.3K]

Answer: 154.09 g

Explanation:

3 Mg + 2 FeCl3 = 3 MgCl2 + 2 Fe

The mass of MgCl2 produced is calculated:

m MgCl2 = 175 g FeCl3 * (1 mol FeCl3 / 162.2 g) * (3 mol MgCl2 / 2 mol FeCl3) * (95.21 g MgCl2 / 1 mol) = 154.09 g

5 0

3 years ago