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serious [3.7K]
2 years ago
14

What mass of calcium carbonate (in grams) can be dissolved by 4.0 g of hcl? (hint: begin by writing a balanced equation for the

reaction between hydrochloric acid and calcium carbonate.)?
Chemistry
2 answers:
stiks02 [169]2 years ago
8 0
The reaction between calcium carbonate and hydrochloric acid can be expressed through the chemical reaction,

    CaCO3 + 2HCl --> CaCl2 + H2O + CO2

The molecular weight of calcium carbonate is 100 g/mol while that of hydrochloric acid is 36.45. The equation above depicts that 100 g of calcium carbonate can be dissolved in 72.9 g of hydrochloric acid. 

    x = (4 g HCl)(100 g CaCO3 / 72.9 HCl)
      x = 5.49 g

Answer: 5.49 g
Paha777 [63]2 years ago
3 0

Answer:

4 grams of HCl need 5.486 grams of CaCO3

Explanation:

The balanced chemical reaction between hydrocloric acid and calcium carbonate can be expressed as

CaCO3+ 2HCl -> CaCl2 + CO2 + H2O

This means that 1 mole of CaCO3 reacts with 2 mole of HCl to produce 1 mole of CaCl2,CO2 and H2O.

2 mole of HCl needs 2 mole of CaCO3  

In term of mass we can write that

72.9 grams of HCl need 100 grams of CaCO3

1 grams of HCl need (100/72.9) grams of CaCO3

4 grams of HCl need ((100*4)/72.9) grams of CaCO3

4 grams of HCl need 5.486 grams of CaCO3

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Olive oil has a density of 0.92g/ml. How much would 1.0 Liter of olive oil weigh in grams?
Julli [10]

Answer:

9.2x10²g

Explanation:

Data obtained from the question include the following:

Density = 0.92g/ml

Volume = 1L = 1 x 1000 = 1000mL

Mass =..?

Density is simply defined as the mass of the substance per unit volume of the substance. Mathematically it can be represented as:

Density = Mass /volume.

Mass = Density x volume

Mass = 0.92 x 1000

Mass = 9.2x10²g.

Therefore, 1L of olive will weigh 9.2x10²g.

5 0
2 years ago
The number of sp2 hybrid orbitals on the carbon atom in CO32– is
Vladimir79 [104]
The number of sp2 hybrid orbitals on the carbon atom in CO32– is 3. Because hybrids = combination of 2 different types of orbitals
sp2 = 1/3 s character + 2/3 p character
7 0
2 years ago
Read 2 more answers
How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?
ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

<em>x</em> = 75/10 = 7.5

5 0
2 years ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container. The carbon dioxide took 1.25 times as lon
Aleks04 [339]

Answer:

CO

Explanation:

From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.

t1/t2=√m1/m2

Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g

1.25/1=√44/m2

Therefore m2=28g CO

7 0
2 years ago
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