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labwork [276]
3 years ago
10

PLEASE ANSWER QUICK

Chemistry
1 answer:
zubka84 [21]3 years ago
3 0

The equation to find the percentage by mass of a solution is:

(Mass of solute/mass of solution) x 100 = percentage by mass

So, let's identify the variables.

Mass of solute: 22g

Mass of solution: 112g

The reason that the solution is 112g and not 90g is because the solution needs to include the mass of the solute and solvent, meaning 90 + 22 is the mass of the solution.

(22/112) x 100 = 19.6%

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Equal volumes of SO2(g) and O2(g) at STP contain the same number of
LuckyWell [14K]

Answer:

Equal volumes of SO2(g) and O2(g) at STP contain the same number of molecules

Explanation:

According to Avogadro Law,

Equal volume of all the gases at same temperature and pressure have equal number of molecules.

This law state that volume and number of moles of gas have direct relation.

When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.

Mathematical relation:

V ∝ n

V/n = K

K is proportionality constant.

When number of moles change from n₁ to n₂ and volume from V₁ to V₂

expression will be,

V₁/n₁ = K     ,     V₂/n₂ = K

V₁/n₁ = V₂/n₂

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3 years ago
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Solve for a, b, c, AND d<br> d<br> C<br> 84°<br> 930<br> 970<br> b
olga55 [171]

Answer:

a = 87

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3 years ago
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If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

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4 years ago
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3 years ago
A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is
Vlada [557]

Answer:

c=0.127\ J/g^{\circ} C

Explanation:

Given that,

Mass of the sample, m = 275 g

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We need to find the specific heat of the metal. The heat required by a metal sample is given by :

Q=mc\Delta T

c is specific heat of the metal

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C

So, the specific heat of metal is 0.127\ J/g^{\circ} C.

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