Write the skeleton equation for the reaction between hydrogen and oxygen that produces water

Which of the statements listed below is negative aspect of a volcanic eruption?

mixas84 [53]

Answer:

C . the dramatic scenery created by volcanic eruptions attracts tourists.

5 0

2 years ago

How is matter and energy conserved when liquid is frozen

Brut [27]

Answer:

The big energy change when water freezes is in the potential energy of interactions between the water molecules.

Explanation:

I am not to sure if this is correct, but I hope it helps in some way.

3 0

3 years ago

Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0L. What temperature must be maintained

Morgarella [4.7K]

Given:

P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?

Assume that the gas ideally, thus, we can use the ideal gas equation:

PV = nRT

where R = 0.0821 L atm/mol K

123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T

solve for T

T = 295.72 K

5 0

3 years ago

Read 2 more answers

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago

The titration of 78.5 mL of an unknown concentration H3PO4 solution requires 134 mL of 0.224 M KOH solution. What is the concent

Katyanochek1 [597]

Answer:

0.127 M.

Explanation:

The balanced equation for the reaction is given below:

H3PO4 + 3KOH —› K3PO4 + 3H2O

From the balanced equation above, we obtained the following data:

Mole ratio of acid, H3PO4 (nA) = 1

Mole ratio of base, KOH (nB) = 3

Data obtained from the question include:

Volume of acid, H3PO4 (Va) = 78.5 mL

Molarity of acid, H3PO4 (Ma) =...?

Volume of base, KOH (Vb) = 134 mL

Molarity of base, KOH (Mb) = 0.224 M

The concentration of the acid, H3PO4 can be obtained as follow:

MaVa / MbVb = nA/nB

Ma x 78.5 / 0.224 x 134 = 1/3

Cross multiply

Ma x 78.5 x 3 = 0.224 x 134 x 1

Divide both side by 78.5 x 3

Ma = (0.224 x 134 x 1) /(78.5 x 3)

Ma = 0.127 M

Therefore, the concentration of the acid, H3PO4 is 0.127 M.

3 0

2 years ago