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3 years ago

What will happen if you add more dilute to a saturated solution

1 answer:

3 0

If you dilute a concentrated solution, the concentration of the obtained solution is smaller than that of the initial solution. But if you add more solute to a concentrated solution, the concentration of the obtained solution is greater than that of the initial.

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Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.

My name is Ann [436]

Acetic acid, HC2H3O2

First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
 Hydrogen:
P1 = ((4)(1)/60)(100%) = 6.67%

 Carbon:
P2 = ((2)(12)/60)(100%) = 40%

Oxygen
P3 =((2)(16) / 60)(100%) = 53.33%

Glucose, C6H12O6

The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 Hydrogen:
P1 = (12/180)(100%) = 6.67%

Carbon:
P2 = ((6)(12) / 180)(100%) = 40%

Oxygen:
P3 = ((6)(16) / 180)(100%) = 53.33%

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.

6 0

3 years ago

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

7 0

3 years ago

What class of elements do iron and nickel belong?

sergey [27]

IM THINKING THEY BELONG IN THIS CLASS OF ELEMENTS VILL B.
Hope i Helped

5 0

3 years ago

Identify the following redox reactions by type. Check all that apply. (a) Fe + H2SO4 → FeSO4 + H2 combination decomposition disp

EastWind [94]

Answer :

(a) displacement reaction

(b) combination reaction

(c) disproportionation reaction

(d) displacement reaction

Explanation :

(a) The given balanced chemical reaction is,

$Fe+H_2SO_4\rightarrow FeSO_4+H_2$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Fe) replace the less reactive element (H).

(b) The given balanced chemical reaction is,

$S+3F_2\rightarrow SF_6$

This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or single product.

(c) The given balanced chemical reaction is,

$2CuCl_2\rightarrow Cu+CuCl_2$

This reaction is a disproportionation reaction in which the chemical species gets oxidized and reduced simultaneously. It is also considered as a redox reaction.

(d) The given balanced chemical reaction is,

$2Ag+PtCl_2\rightarrow 2AgCl+Pt$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Ag) replace the less reactive element (Pt).

7 0

3 years ago

The camel stores the fat tristearin (C57H110O6) in its hump. As well as being a source of energy, the fat is also a source of wa

Shalnov [3]

Answer:

8.1 × 10² g

Explanation:

Step 1: Write the balanced equation

2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O

Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g

We will use the conversion factor 1 lb = 453.592 g.

1.6 lb × 453.592 g/1 lb = 7.3 × 10² g

Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆

The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.

7.3 × 10² g × 1 mol/890.83 g = 0.82 mol

Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆

The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol

Step 5: Calculate the mass corresponding to 45 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

45 mol × 18.02 g/mol = 8.1 × 10² g

4 0

3 years ago