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2 years ago

How many grams of Ag2CO3 are required to reacr with 28.5 mL of 1.00 M NaOH solution?

1 answer:

8 0

The reaction between Ag2CO3 and NaOH is shown by the equation below
Ag2CO3 + NaOH = Ag2O + Na2CO3 +H2O
we can determine the number of mole of sodium hydroxide
by (2.85 ml × 1) ÷ 1000 ml , since according to molarity 1 mole is contained in 100ml.
we get 0.00285 moles of NaOH
Using the mole ratio we can get the moles of Ag2CO3
Mole ratio: Ag2NO3 : NaOH = 1:1
Therefore, the moles of Ag2CO3 will be 0.00285 moles
but 1 mole of silver carbonate is equivalent to 275.8 g
Thus the mass will be calculated by 0.00285 moles × 275.8g = 0.78603g
Mass of silver carbonate required will be 0.78603g

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A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu

tekilochka [14]

Answer:

$x_B=0.0769$

$m=0.990m$

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

$x_B=\frac{n_B}{n_B+n_C}$

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

$n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol$

Thus, we compute the mole fraction:

$x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769$

Next, for the molality, we define it as:

$m=\frac{n_B}{m_C}$

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

$m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg$

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0

2 years ago

A gas is contained in a thick-walled balloon. When the pressure changes from 319 mm Hg to 215 mm Hg, th volume changes from 0.55

Dmitry_Shevchenko [17]

In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L

7 0

2 years ago

Read 2 more answers

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. Th

Setler [38]

Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, 1/s' + 1/s = 1/f

2. magnification = h'/h = s'/s

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4 = 120cm

s' = 1.4*120 = 160cm

Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens

Since the object is at a distance between f and 2f from the lens, the image formed is real, inverted and magnified

3 0

2 years 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

2 years ago

A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat o

Radda [10]

Answer:

The specific heat for the titanium metal is 0.524 J/g°C.

Explanation:

Given,

Q = 1.68 kJ = 1680 Joules

mass = 126 grams

T₁ = 20°C

T₂ = 45.4°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

Here, ΔT = T₂ - T₁ = 45.4 - 20 = 25.4°C.

Substituting values,

1680 = (126)(25.4)(Cp)

By solving,

Cp = 0.524 J/g°C.

The specific heat for the titanium metal is 0.524 J/g°C.

3 0

2 years ago