Help please I’m confused

The gravitational pull of the Moon has the greatest influence on the water levels of Earth’s ocean tides. If the distance betwee

kherson [118]

The water levels would lower, because as the moon gets farther away, it's gravitational pull decreases. Hope this helps! :)

6 0

2 years ago

Write a balanced net ionic equation for the following reaction. BaCI2(aq)+H2SO4(aq) -> BaSO4(s)+HCI (aq)

Sever21 [200]

Ba²⁺ + 2Cl⁻ + 2H⁺ + SO₄²⁻ = BaSO₄ (precipitate) + 2H⁺ + 2Cl⁻
Ba²⁺ + SO₄²⁻ = BaSO₄

4 0

3 years ago

Read 2 more answers

How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0

2 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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3 0

1 year ago

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammo

Kryger [21]

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

8 0

3 years ago