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3 years ago

6th grade help me pleaseeeee

Chemistry

1 answer:

Vesnalui [34]3 years ago

7 0

Answer:

the answer would be A. cells

Explanation:

just trust me Im a 8th grader and I did that before

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If the mass of a material is 78 grams and the volume of the material is 23 cm3, what would the density of the material be?

Paraphin [41]

Density = mass / volume

Density = (78g) / (23cm^3)

Density = 3.4g/cm^3

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3 years ago

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What do acids do in solution?

KiRa [710]

B
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2 years ago

Do all waves need a material to travel through? HELPPPP

Natasha2012 [34]

No.
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hope this helps and have a great day :)

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3 years ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

8 0

2 years ago

Read 2 more answers

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago