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hodyreva [135]
3 years ago
7

Which factor has the greatest effect on object X remaining in its orbit?

Chemistry
1 answer:
saw5 [17]3 years ago
3 0

Explanation:

Stop cheating and use your brains kiddos

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11.
jolli1 [7]

Answer:

Number of protons and neutrons

7 0
3 years ago
what is the difference between the number of electrons in an atom of selenium and the number of electrons in an atom of aluminum
igomit [66]

Answer:

The atomic number of Selenium is 34. This means that Selenium possesses 34 electrons.

The atomic number of Aluminium is 13. This means that Aluminium has 13 electrons.

Hence, there is a difference of 21 between the number of electrons in an atom of selenium and the number of electrons in an atom of aluminium.

Selenium has 6 electrons in it's outer most shell whereas aluminium has 3 electrons in its outer most shell. As a result, aluminium will have a greater tendency to lose one of its outer most electrons to become stable.

7 0
3 years ago
Select the correct answer. if two half-lives have passed since a scientist collected a 1.00-gram sample of u-235, how much u-235
Ugo [173]

Answer:

0.25 g of U-235 isotope will left .

Formula used :

where,

N = amount of U-235 left after n-half lives = ?

= Initial amount of the U-235 = 1.00 g

n = number of half lives passed = 2

0.25 g of U-235 isotope will left .

3 0
2 years ago
How many moles are in 39.5 grams of Lithium?
Blizzard [7]

Answer:

185.05 g.

Explanation

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2LiCl → 2Li + Cl₂

It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.

We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.

n = (30.3 g) / (6.941 g/mole) = 4.365 moles.

Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.

Using cross multiplication:

2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.

??? moles of LiCl → 4.365  moles of Li.

The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.

Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).

Molar mass of LiCl = 42.394 g/mole.

mass = n x molar mass = (4.365 x 42.394) = 185.05 g.

7 0
3 years ago
Read 2 more answers
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

[HNO_3] = 0.200 M

[Ca(OH)_2] =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

6 0
4 years ago
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