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mamaluj [8]
2 years ago
12

The recommended daily allowance (rda of calcium is 1.2 g. calcium carbonate contains 12.0% calcium by mass. how many grams of ca

lcium carbonate are needed to provide the rda of calcium?
Chemistry
2 answers:
Y_Kistochka [10]2 years ago
6 0

Answer:

10 grams of calcium carbonate are needed to provide the recommended daily allowance of calcium.

Explanation:

The recommended daily allowance (rda) of calcium = 1.2 g

Percentage of calcium by mass in calcium carbonate = 12.0%

Total mass of calcium carbonate be x

\%=\frac{\text{Mass of an element}}{\text{Total mass of compound}}\times 100

12.0\%=\frac{1.2 g}{x}\times 100

x = 10 g

10 grams of calcium carbonate are needed to provide the rda of calcium.

defon2 years ago
4 0
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!

2) Mass fraction of this is excessive data.

3) The solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
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How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu
Alexxandr [17]
The  moles  of  chromium (iii)  nitrate  produced  is  calculated  as   follows

write  the  equation  for  reaction

 3  Pb(NO3)2  +  2 Cr  =  2 Cr(NO3)3  +  3  Pb

by  use  of  mole  ratio  between  Pb(NO3)2  to  Cr(NO3)3  which  is  3  :  2  the  moles  of  Cr(NO3)3  is therefore  
=  0.85  x2  /3  =  0.57   moles
5 0
2 years ago
Stabiliţi numerele de oxidare ale tuturor elementelor prezente în următoarele substanţe chimice, ţinând cont de principalele reg
natita [175]

Answer:

a.

N = +2

O = -2

b.

Na = +1

O = -2

N = +5

c.

O = -2

Al = +3

P = +5

d.

O = -2

Ca = +2

C = +4

e

O = -2

Mn = +7

K = +1

f.

O = -2

K = +1

Cr = +6

g.

O = -2

Cl = +7

Cr = +1

h.

O = -2

H = +1

Cr = +5

i.

O = -2

H = +1

Cr = +3

k.

O = -2

H = +1

Cr = +1

Explanation:

A. NU

Numărul de oxidare a azotului = +2

Numărul de oxidare a oxigenului = -2

b. NaNO₃

Numărul de oxidare de Na = +1

Numărul de oxidare de O = -2

Numărul de oxidare de N = +5

c. AlPO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Al = +3

Numărul de oxidare al P = +5

d. carbonat de calciu

Numărul de oxidare de O = -2

Numărul de oxidare de Ca = +2

Numărul de oxidare de C = +4

e. KMnO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Mn = +7

Numărul de oxidare al lui K = +1

f. K₂Cr₂O₇

Numărul de oxidare de O = -2

Numărul de oxidare al lui K = +1

Numărul de oxidare al Cr = +6

g. HClO₄

Numărul de oxidare de O = -2

Numărul de oxidare al Cl = +7

Numărul de oxidare al Cr = +1

h. HClO₃

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +5

i. HClO₂

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +3

k. HClO

Numărul de oxidare de O = -2

Numărul de oxidare al lui H = +1

Numărul de oxidare al Cr = +1.

6 0
1 year ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
1 year ago
What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercu
Olegator [25]

Answer:

\Delta T=194^oC

Explanation:

Hello,

In this case, in terms of the heat, mass, heat capacity and change in temperature, we can analyze thermal changes as:

Q=mCp\Delta T\\

In such a way, we compute the required change in temperature as shown below:

\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033\frac{cal}{g^oC} }  \\\\\Delta T=194^oC

Such change in temperature is positive indicating an increase in the temperature as the involved heat is positive, in means that heat was added to increase the temperature.

Best regards.

6 0
1 year ago
Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressur
Gnoma [55]

Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

t_{1/2}= \frac{0.693}{K}

So, fraction of original pressure = \frac{1}{2}^2

n here is number of half life

therefore, \frac{1}{8}= \frac{1}{2}^3

⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

8 0
2 years ago
Read 2 more answers
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