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3 years ago

8.3 kg, 1.7 Kg/cm3 -> ? cm3

Chemistry

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

4.9 cm³

Explanation:

Given data:

Mass value = 8.3 kg

Density = 1.7 kg/cm³

Volume = ?

Solution;

In given question from the units we got to know that the mass and density value is given and we have to calculate the volume.

Thus, we will apply density formula.

Density = mass/ volume

d = m/v

1.7 kg/cm³ = 8.3 kg/ v

v = 8.3 kg/ 1.7 kg/cm³

v = 4.9 cm³

Thus, the volume is 4.9 cm³.

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When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction oc

Georgia [21]

Answer:

$2Cl^{-}_{(aq)} + 2K^{-} _{(aq)} = 2KClx_{(aq)}$

Explanation:

The reaction is a precipitation reaction. In other words, the two aqueous solutions react to give the solid salt, potassium chloride (KCl) in this case.

$SrCl_{2} _{(aq)} + K_{2}SO_{4} _{(s)} = SrSO_{4} _{(aq)} + 2KCl_{(s)}$

Because the reaction product is a solid, the net ionic equation can be written as:

$2Cl^{-}_{(aq)} + 2K^{-} _{(aq)} = 2KClx_{(aq)}$

This is the resultant equation after removing spectator ions: sulfate and strontium.

3 0

3 years ago

When the fish-tank water has a pH of 8.0, the hydronium ion concentration is 1.0 × 10-8 mole per liter. What is the hydronium io

Oduvanchick [21]

You must know and use the formula for pH.

pH = - log [H3O+], where [H3O+] is the molar concentration of hydronium ion.

So, when pH is 8.0 => 8.0 = - log [H3O+] and you can use antilogarithm (the inverse function of logarithm) to find [H3O+], in this way:

[H3O+] = 10^-8 = 1 * 10 ^-8 M

When, pH = 7.0 =>

7.0 = - log [H3O+] => [H3O+] = 1 * 10^ -7 M

Answer: 1*10^-7 mole / liter

5 0

3 years ago

A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma

-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we use PV=nRT to calculate n, which is the total number of moles of nitrogen and hydrogen:

1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K

n = 0.307 mol

So now we know that

MolH₂ + MolN₂ = 0.307 mol

and

MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a system of two equations and two unknowns. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

MolH₂ + MolN₂ = 0.307 mol

MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49

(0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49

0.614 - 2MolN₂ + 28molN₂ = 3.49

0.614 + 26MolN₂ = 3.49

MolN₂ = 0.111 mol

Now we calculate MolH₂:

MolH₂ + MolN₂ = 0.307 mol

MolH₂ + 0.111 = 0.307

MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to calculate the mass percent:

N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂

H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0

3 years ago

Suppose that you are performing a titration on a monoprotic acid. Titration of this monoprotic acid required a volume of 0.1L of

Alex73 [517]

Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions

Explanation:

The monoprotic acid (HA) has a valency of 1 and diprotic acid $(H_2A)$ has a valency of 2.

As the concentration and volume of the diprotic acid and the monoprotic acids are equal.

The neutralization reaction for monoprotic acid is:

$HA+BOH\rightarrow BA+H_2O$

The neutralization reaction for diprotic acid is:

$H_2A+2BOH\rightarrow B_2A+2H_2O$

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.

6 0

3 years ago

PLEASE ANSWER 30 POINTS !!

Alexeev081 [22]

Can u explain it more plz.

3 0

3 years ago

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