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creativ13 [48]
3 years ago
11

Which ion in each of the following pairs would you expect to be more strongly hydrated Why?ClO4- or SO4-

Chemistry
1 answer:
Viktor [21]3 years ago
3 0

Answer:

ClO₄⁻

Explanation:

When an ion is hydrated it is surrounded by water molecules, thus, as small is the ion, more molecules may surround it, and it will be more strongly hydrated. In this case, the Cl is small than the S atom, because Cl is from group 17, and S from group 16, and Cl has more valence electrons, which will be more attracted to the nuclei.

So, ClO₄⁻ will be more strongly hydrated.

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What kind of solid is crystalline boron
Ulleksa [173]

Answer:

It is a network solid, a lattice of many covalent bonds (like diamond, except that it is black rather than transparent).

Explanation:

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3 years ago
Do plants have to do cellular respiration?<br> Why or why not?
kozerog [31]

Answer:

<h3>Plants need oxygen because plants are made living cells which need to perform cellular respiration to get the energy they need to function. Admittedly, needing to perform cellular respiration alone does not mean that consuming oxygen is required - many (micro)organisms use fermentation or anaerobic respiration.</h3>
8 0
1 year ago
How many grams of water can be produced from 10 moles of hydrogen gas?
Veseljchak [2.6K]
10 moles of H2 reacts with 5 moles of O2 creating 10 moles of water.
Water has the molar mass 18 g/mole
10 · 18g = 180 g 
180 g of water can be produced from 10 moles of hydrogen gas.
8 0
4 years ago
Jane is sliding down a slide. What kind of motion is she demonstrating?
Marta_Voda [28]
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4 0
3 years ago
The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

4 0
3 years ago
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