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almond37 [142]
4 years ago
7

Is magnesium carbonate soluble in water

Chemistry
2 answers:
KiRa [710]4 years ago
8 0

Magnesium carbonate can do this because it <u>doesn't dissolve in water,</u> only acid, where it will effervesce (bubble).

seropon [69]4 years ago
5 0

Answer:

no it is not, it is only soluable in acid.

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A cup of gold colored metal beads was measured to have a mass 425 grams. By water displacement, the volume of the beads was calc
Troyanec [42]

Given parameters;

Mass of gold colored metal beads  = 425g

Volume of water displaced by beads = 48.0cm³

Unknown;

Identity of the metal = ?

Given densities;

Gold: 19.3 g/mL

Copper: 8.86 g/mL

Bronze: 9.87 g/mL

Density is an intensive property of any substance. This implies that we can use the density of any substance to identify it.

Density can be defined as the mass per unit volume of a substance. Every substance has a unique mass per volume.

 Mathematically;

                    Density  = \frac{mass}{volume}

where mass is in kg or g

           volume is in  m³ or cm³

To find the density, we must know the mass and volume.

In this problem, the volume of the gold metal beads is the same as the volume of water displaced. This is a way to measure volume of solids.

Since the volume is given in cm³, and we are comparing with choices that have units in g/mL, we simply convert the volume in cm³ to mL

                   1cm³  = 1mL³

So therefore, volume of gold colored metal is 48mL

Now input the parameters given and solve for the density;

           Density  = \frac{425g}{48mL }    = 8.85g/mL

From the given densities, we clearly see that copper is the metal since they both of similar densities.

4 0
3 years ago
which organelle converts the chemical energy stored in food into compounds that are more convenient for the cell to use?
stiv31 [10]
It's the mitochondrion. It it the powerhouse of the cell and deals mainly with energy.
6 0
4 years ago
Read 2 more answers
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg
Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 0
3 years ago
An ideal diatomic gas at 80 k is slowly compressed adiabatically and reversibly to half its volume. what is its final temperatur
Yanka [14]

An ideal diatomic gas at 80 k is slowly compressed adiabatically and reversibly to half its volume. The final temperature is 104.8k

<h3>Calculation of final temperature </h3>

The formula used for compression is:-

TV^(γ-1)=C

where;

T= temperature=80k

V=volume(given volume is half of its original volume i.e v/2)

γ=CP/CV= ( 7.R/2)/(5R/2)=7/5

C= constant

using  the given  values in the formula;

80 x V^[(7/5)-1]=T(final) (V/2)^[(7/5)-1]

80=T(final) x (V/V2)^(2/5)

80=T(final)  x (1/2)^(2/5)

T(final)= 80 x  (2)^2/5

T(final)= 80 x 1.31

final temperature  =104.8k

Learn more about Temperature here:-

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6 0
1 year ago
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

7 0
3 years ago
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