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Licemer1 [7]
3 years ago
13

PLZ HELP WITH THESE QUICK// WILL AWARD BRANLIEST

Chemistry
1 answer:
Tom [10]3 years ago
5 0

Answer 1

The sample will dissolve in more than 1 minute.

Explanation :-

Generally Solubility and rate of solubility of substances increase with the increase in temperature. So lower the temperature less fast will it dissolve in the same amount of water.

Since at 50 C it takes 1 minute to dissolve, at 20 C which is lower temperature it will take more time to dissolve.

Answer 2

Transition metals

Explanation:-

Hardness depends on the extent of metallic bonding for metals. More the number of electrons more the metallic bonding.

Alkali metals with just 1 valence electron have weak metallic bonding. Alkaline earth metals have just 2 valence electrons. Transition elements has more electrons in penultimate shell and valence shell than lanthanides. Transition metals with most metallic bonding are the hardest

You might be interested in
A chemist observes that a large molecule reacts as if it were much smaller. The chemist proposes that the molecule is folded in
nadezda [96]

Answer:

Hypothesis

Explanation:

The following steps are applicable when we wish to prove a specific fact:

  • a hypothesis is made; this is a statement that we provide after some observations and we wish to either prove or deny it;
  • multiple experiments are carried out in order to gather significantly substantial amount of data that can be then further analyzed and any tendencies can be noticed;
  • based on the data gathered, conclusions are made: we either prove or deny the hypothesis. If hypothesis is proved, it may become a theory over long time.

In the context of this problem, we're at the first step where we make a hypothesis.

6 0
3 years ago
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O Wh
tamaranim1 [39]

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

4 0
3 years ago
How do you do these?
Agata [3.3K]

Solve these problems like weighted averages:

The first one:

Multiply the masses (isotope numbers) by the decimal form of the percentage. Add them

0.076 (6) + 0.924 (7) = 6.924


The second one:

0.2 (10) + 0.8 (11) = 10.8


If you think about it, these answers make sense. 6.924 is much closer to 7 than to 6 (since there's a lot more lithium-7 than there is lithium-6). 10.8 is closer to 11 than to 10.


6 0
3 years ago
Which type of wave cannot move without a medium in which to travel?
Zanzabum

sound wave cannot travel without a medium.

5 0
3 years ago
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