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3 years ago

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Chemistry

1 answer:

allsm [11]3 years ago

4 0

Answer:

% Mass of sand = 35.95%

% Mass of KBr = 64.05%

Explanation:

From the question given, we obtained the following:

Mass of mixture( KBr + sand) = 10g

Mass of sand = 3.595g

Mass of KBr = 10 — 3.595 = 6.405g

% Mass of sand = ( Mass of sand / mass of mixture) x 100

% Mass of sand = (3.595 / 10) x100

% Mass of sand = 35.95%

We can obtain the percentage of KBr by subtracting the percentage of sand from 100. This is illustrated below:

% Mass of KBr = 100 — 35.95

% Mass of KBr = 64.05%

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As a factor of production, land best refers to

Maru [420]

The first factor of production is land, but this includes any natural resource used to produce goods and services. This includes not just land, but anything that comes from the land. Some common land or natural resources are water, oil, copper, natural gas, coal, and forests.

So I’m assuming the answer would be:

C. All natural resources

5 0

3 years ago

Read 2 more answers

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

cluponka [151]

<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>

3 0

3 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

A chemist adds 485 mL of a 0.0025 mol/L calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sul

viva [34]

Answer : The mass in grams of calcium sulfate is 0.16 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one litre of solution.

Formula used :

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}$

Solute is, $CaSO_4$

Given:

Molarity of $CaSO_4$ = 0.0025 mol/L

Molar mass of $CaSO_4$ = 136 g/mole

Volume of solution = 485 mL

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

$0.0025=\frac{\text{Mass of }CaSO_4\times 1000}{136\times 485}$

$\text{Mass of }CaSO_4=0.16g$

Thus, the mass in grams of calcium sulfate is 0.16 grams.

6 0

3 years ago

The expression below was formed by combining different gas laws. V is proportional to StartFraction n T over P EndFraction. Whic

Ronch [10]

Answer:

The Ideal gas law

Explanation:

From the given question, we have:

V $\alpha$ $\frac{nT}{P}$

where each variable has its usual meaning.

Thus,

V = $\frac{nRT}{P}$

where R is the ideal gas constant

cross multiply to have;

PV = nRT

This implies that the volume of the gas is directly proportional to the number of moles of the gas.

Therefore, the law can be used to determine the relationship between the volume and number of moles is the ideal gas law.

3 0

3 years ago