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3 years ago

PLZ HELP! WILL GIVE BRAINLIEST, 5 STARS, AND A THANKS!!

Chemistry

1 answer:

nika2105 [10]3 years ago

8 0

Answer:

The similarities between freezing, and condensation is, the stimulis.

Explanation:

The explanation to that is because the condensation will melt.

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What is the formula of titanium(II) oxide?(1) TiO (3) Ti2O(2) TiO2 (4) Ti2O3

ddd [48]

Hehe I love Titanium!! ;) Anyways, the chemical formula of titanium (II) Oxide is TiO

6 0

3 years ago

A 5.00 gram sample of magnesium sulfate hydrate (epsom salt) is heated in the lab to form anhydrous magnesium sulfate. After hea

Ostrovityanka [42]

Answer:

The answer to your question is MgSO₄ 5H₂O

Explanation:

Data

mass of MgSO₄ = 2.86 g

mass of H₂O = 2.14 g (5 - 2.86)

Process

1.- Calculate the molecular mass of the compounds

MgSO₄ = 24 + 32 + (16 x 4) = 120

H₂O = 16 + 2 = 18

2.- Convert the grams obtain to moles

120 g of MgSO₄ --------------- 1 mol

2.8 g ---------------- x

x = (2.8 x 1)/120

x = 0.024 moles

18 g of H₂O --------------------- 1 mol

2.14 g -------------------- x

x = (2.14 x 1)/18

x = 0.119

3.- Divide by the lowest number of moles

MgSO₄ = 0.024/0.024 = 1

H₂O = 0.119/ 0.024 = 5

4.- Write the molecular formula

MgSO₄5H₂O

5 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Cl2(g) + 2kBr(s) ---> 2KCl(s) + Br2(g) Rewrite the equation and write the color of each chemical under its name. 2. What type

kolezko [41]

Answer:

Cl2(g) (green/yellow mix) + 2KBr(s) (white) ---> 2KCl(s) (violet) + Br2(g) (reddish brown)

This chemical reaction is a redox type.

Explanation:

Look at the oxidation state, when the number increase your element gets oxidated, when the number decrease, the elements it's getting reduced.

8 0

2 years ago

An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j

nydimaria [60]

Answer:

Cp = 0.237 J.g⁻¹.°C⁻¹

Explanation:

Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 640 J

m = mass = 125 g

Cp = Specific Heat Capacity = <u>??</u>

ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 640 J / (125 g × 21.6 °C)

Cp = 0.237 J.g⁻¹.°C⁻¹

3 0

3 years ago