Determine the mass, in grams, of 0.350 moles of s (1 mol of s has a mass of 32.07 g).

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

2 years ago

If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye

sladkih [1.3K]

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

Energy of light (E): 2.79 × 10⁻¹⁹ J

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Speed of light (c): 3.00 × 10⁸ m/s

Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

5 0

2 years ago

Please help me will mark brainliest

LuckyWell [14K]

Answer:

accretion

Explanation:

the coming together and cohesion of matter under the influence of gravitation to form larger bodies.

8 0

2 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$