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3 years ago

When does surface water become groundwater?

Chemistry

1 answer:

Sidana [21]3 years ago

8 0

Answer:

When it rains or snows. Basically as soon as it hits the ground.

Explanation:

Groundwater begins as rain or snow that falls to the ground. This is called precipitation but only a small portion of this precipitation will become groundwater

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What is the ph of a solution that results from mixing 25 ml of 0.15 M hcl to 25 ml of 0.52 m nh3?

BartSMP [9]

Answer:

7.00

Explanation:

When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.

First, let's find the number of moles of each one of them, multiplying the concentration by the volume:

nH+ = 0.15 M * 25 mL = 3.75 mmol

nNH3 = 0.52 M * 25 mL = 13 mmol

So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.

5 0

3 years ago

How many moles of LiOH are needed to react completely with 25.5 g of CO2

LekaFEV [45]

Answer:

3.18 mol

Explanation:

$2LiOH+CO_{2}-> Li_{2}CO_{3} +H_{2}O$

n(CO2) = mass/ Mr.

= 25.5 / 16

= 1.59 mol

As per the equation above,

n(LiOH) : n(CO2)

2 : 1

∴ 3.18 : 1.59

3 0

3 years ago

2. Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP, is needed to com

Juliette [100K]

Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)

when we have STP conditions, we can use this conversion: 1 mol = 22.4 L

first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.

molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)

calculations:

$6.25 g Na ( \frac{1 mol Na}{23.0 g} ) ( \frac{1 mol Cl_2}{2 mol Na} ) ( \frac{22.4 L}{1 mol} )= 3.04 L$

8 0

3 years ago

Water is Carried to the leaves by stomata

tiny-mole [99]

Answer:

please explain further and i maybe can help you

Explanation:

5 0

3 years ago

Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).

nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE for C≡O = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

brainly.com/question/13981382

#SPJ1

7 0

1 year ago