What are the types of precipitation?

A solution has a concetration of 0.3mol/dm3 of sodium hydroxide .what volume of

HACTEHA [7]

Answer:

Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O

1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M

Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol

7 0

3 years ago

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

Eva8 [605]

Answer:

$[A]_0=0.400M$

Explanation:

Hello.

In this case, since the first-order reaction is said to be linearly related to the rate of reaction:

$r=-k[A]$

Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:

$\frac{d[A]}{dt} =-k[A]$

Which integrated is:

$ln(\frac{[A]}{[A]_0} )=-kt$

And we can calculate the initial concentration of the hydrogen peroxide as follows:

$[A]_0=\frac{[A]}{exp(-kt)}$

Thus, for the given data, we obtain:

$[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}$

$[A]_0=0.400M$

Best regards!

3 0

3 years ago

Find the percent composition of C7H5N3O6

Oksana_A [137]

Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.

First, let's find the total mass by using the masses of the elements given on the periodic table.

7 x 12.011 (mass of Carbon) = 84.077

5 x 1.008 (mass of Hydrogen) = 5.04

3 x 14.007 (mass of Nitrogen) = 42.021

6 x 15.999 (mass of Oxygen) = 95.994

Add all of those pieces together.

84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.

Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %

Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %

Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %

Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %

You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.

5 0

3 years ago

How much would a 10 kg suitcase weigh on the surface of…? a. The Moon b. Mars c. Saturn d. Pluto

alukav5142 [94]

Answer:

Is this the full question

Explanation:

Reply or love the message so i can get a notification

8 0

2 years ago

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago