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4 years ago

Answer the following question.

Chemistry

1 answer:

VMariaS [17]4 years ago

4 0

Answer:

A. Na2SO4 + BaCl2 ----> BaSO4(solid, white) +2NaCl doublr replacement reaction

B. MnO2 + 4HCl ----> MnCl2 + 2H2O + Cl2 (gas) decomposition

C. CuSO4 + Fe ---> FeSO4 + Cu single replacement reaction

You might be interested in

How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex

Brut [27]

When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M
so by substitution we can get [Ca2+]
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
= 0.00167 mol

4 0

4 years ago

A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.

makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> smallest whole number ratio of elements in a compound

Molecular Formula => actual whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

% => grams => moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H => 7.14 : 14.29

To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0

3 years ago

If 25.3 grams of mercury(II) oxide react to form 23.4 grams of mercury, how many grams of oxygen must simultaneously be formed?

sergij07 [2.7K]

Answer : The mass of oxygen formed must be 3.8 grams.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$2HgO\rightarrow 2Hg+O_2$

According to the law of conservation of mass,

Total mass of reactant side = Total mass of product side

Total mass of $2HgO$ = Total mass of $2Hg+O_2$

As we are given :

The mass of $HgO$ = 25.3 grams

The mass of $Hg$ = 23.4 grams

So,

$2\times 25.3g=2\times 23.4g+\text{Mass of }O_2$

$50.6g=46.8g+\text{Mass of }O_2$

$\text{Mass of }O_2=3.8g$

Therefore, the mass of oxygen formed must be 3.8 grams.

6 0

3 years ago

An 80 kg long jumper in flight accelerates at a rate of 10 m/s^2. What is the force of the long jumper

Katarina [22]

Answer:

F = 800 N

Explanation:

Given data:

Mass = 80 Kg

Acceleration = 10 m/s²

Force = ?

Solution:

Formula:

F = m × a

F = force

m = mass

a = acceleration

Now we will put the values in formula:

F = m × a

F = 80 kg × 10 m/s²

F = 800 kg.m/s²

kg.m/s² = N

F = 800 N

6 0

3 years ago

4. A compact disc has a diameter of 11.8 cm. What is the surface area of the disc in square centimeters? In square meters? [Area

d1i1m1o1n [39]

The surface area is 109.3 square centimeters or 0.01093 square meters. The area formula requires that we use the radius of the disc. We can find the radius by diving the diameter by 2, so radius = 11.8/2 or 5.9 cm. We can use 3.14 as an approximation for Ď€. The surface area is 3.14 * (5.9*5.9). Since the diameter is given in cm, the surface area units are in square centimeters. To convert to meters, divide any measurement in centimeters by 100, but we need to convert to "square" meters, so we need to divide our square centimeters by 100 * 100, or by 10,000. Dividing 109.3 by 10,000 results in 0.01093 square "meters".

8 0

4 years ago