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Klio2033 [76]
2 years ago
10

describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid​

Chemistry
1 answer:
Reptile [31]2 years ago
5 0

Answer:

Explanation:

250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.

150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.

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How many moles are in 1.2x10^3 grams of ammonia(NH3) ?
miskamm [114]

Answer : The number of moles present in ammonia is, 70.459 moles.

Solution : Given,

Mass of ammonia = 1.2\times 10^3g

Molar mass of ammonia = 17.031 g/mole

Formula used :

\text{Moles of }NH_3=\frac{\text{ given mass of }NH_3}{\text{ molar mass of }NH_3}

\text{Moles of }NH_3=\frac{1.2\times 10^3g}{17.031g/mole}=70.459moles

Therefore, the number of moles present in ammonia is, 70.459 moles.

5 0
3 years ago
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Jim and Jan were racing in a 400 meter event. Jim ran the race in 44.32 seconds. Jan ran the race in 42.21 seconds. Calculate th
nordsb [41]

Answer:

2.11

Explanation:

44.32- 42.21=2.11

Hope this helps

4 0
2 years ago
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What holds the two strands of a DNA molecule together?A. Ionic bonds between negatively and positively charged ionsB. Covalent b
statuscvo [17]

Answer:

D. Hydrogen bonds between complementary base pairs89

Explanation:

A DNA molecule is composed of two long polynucleotide chains made of four types of nucleotide subunits, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). These nucleotides are joined by covalent bonds forming a phosphate-sugar backbone. <em>These strands are held to one another with hydrogen bonds between the base portions of complementary nucleotides.</em>

I hope you find this information useful and interesting! Good luck!

8 0
3 years ago
A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o
HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Therefore, the concentration of the solution is 1.1 gmL¯¹

3 0
3 years ago
When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, O_2

Solution : Given,

Moles of FeCl_3 = 3.00 mole

Moles of O_2 = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)

From the balanced reaction we conclude that

As, 2 moles of FeCl_3 react with 1 mole of O_2

So, 3.00 moles of FeCl_3 react with \frac{3.00}{2}=1.5 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and FeCl_3 is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, O_2

4 0
3 years ago
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