Suppose you want to make an acetic acid/acetate buffer to a pH of 5.00 using 10.0 mL of 1.00 M acetic acid solution. How many milliliters of 1.00 M sodium acetate solution would you need to add? The pKa for acetate buffer is 4.75.

Answer: There are 4.375 moles in 2.5 L of 1.75 M [tex]Na_2CO_3[/tex]

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]      

Molarity of solution = 1.75 M

Volume of solution = 2.5 L  

Putting values in equation , we get:

[tex]1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol[/tex]

Answer: 248.66g

Explanation:

4.3e24 / 6.23e23 = 6.9 mols O2

6.9(18.02*2)=248.66

Answer:

True

Explanation:

yes magnetic feild lines is always combined

Answer:

correct formula: C2F4

Explanation:

tetra means 4. There are 2 C and 4 F in dicarbon tetrafluoride.

Answer:

stars and lightning. this is the answer ok

Answer:

The answer is C. six months

Explanation:

I hope this helps you bestie:))

Answer:

[tex]\% N=82.2\% \\\\\% H=17.8\%[/tex]

Explanation:

Hello there!

In this case, since ammonia is NH₃, we can see nitrogen weights 14.01 g/mol and hydrogen 3.03 g/mol as there is one only nitrogen atom and three hydrogen atoms; thus, the total mass is 17.04 g/mol. In such a way, the percent composition of each element turns out to be:

[tex]\% N=\frac{14.01}{17.04}*100\%=82.2\% \\\\\% H=\frac{3.03}{17.04}*100\%=17.8\%[/tex]

Best regards!

Answer:

Amplitude

Explanation:

Hope I helped :)

A severe storm would be most likely to damage a coastal area by causing pollution. Therefore, the correct option is option B.

The region where land joins the ocean, or even as a line defining the border between the land as well as the shoreline, is known as the coast, sometimes referred as the coastline and seashore. The terrain of the area's surroundings has an impact on shorelines, as do water-induced erosion processes like wave action.

The length of the shoreline on Earth is about 620,000 kilometers. Natural ecosystems' coasts are significant zones and frequently the site of a diverse range of biodiversity. They are home to significant ecosystems on land, such as freshwater and estuarine wetlands, that are crucial for the survival of bird populations or other land animals. A severe storm would be most likely to damage a coastal area by causing pollution.

Therefore, the correct option is option B.

To know more about coastal area, here:

https://brainly.com/question/14105067

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Answer:

From 00:11

Balance the Molecular Equation

From 01:35

Ions for the Complete Ionic Equation

From 03:04

Net Ionic EquationExplanation:

the answer is A I took the test

Answer:

Ca + 2 HCl -- CaCl2 + H2

Explanation:

There are two hydrogen on the right side, so we multiply the left side( HCl) by 2 to get 2 hydrogen on the left. This automatically gives you 2 Cl on left side and there are already 2 cl on the right side, so now this equation is balanced.

Answer:

c.

air molecules are close together and pressing on Earth’s surface.

Explanation:

i hope this helps!

Answer:

53.3g/mol

Explanation

Moles=mass/molar mass

Molar mass= mass/moles

Molar mass= 192/3.6

= 53.3

Answer:

John Dalton

Explanation:

Both John Dalton and Democritus thought that the atom was an indivisible sphere until J.J. Thompson came out with the plum pudding model. Hope I helped!

Answer:

2Fe + 3Cl2 → 2FeCl3

Explanation:

Answer:

V₂ = 25.065 L

Explanation:

Given data:

Initial volume = 15L

Initial pressure = 1500 mmHg (1500/760 = 1.97 atm)

Initial temperature = 299 K

Final temperature = 350 K

Final volume = ?

Final pressure = 1050 mmHg (1050/760 = 1.38 atm)

Formula:

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1.97 atm × 15 L × 350 K / 299 K × 1.38 atm

V₂ = 10342.5 atm .L. K / 412.62 K.atm

V₂ = 25.065 L

Answer:

[tex]0.8\overline 3[/tex] moles of A and 2.5 moles of B in the one-phase system

[tex]1.1 \overline 6[/tex] moles of A and [tex]0.1458 \overline 3[/tex] moles of B in the two-phase system

[tex]0.3541 \overline 6[/tex] moles of B remains in the system

Explanation:

The given parameters are;

The extent of miscibility of liquid B and C = Partially miscible

The number of moles of liquid B added to 1 mole of liquid A that forms a two-phase system = 0.125 mol of liquid B

The number of moles of liquid B added to 1 mole of liquid A that forms a one -phase system = 3 moles of liquid B

Whereby the system consist of 2.5 mol of B and 2 mol of C at 25°C, we have;

The 1 mole of A mixes with 3 moles of B to form a single phase solution

Therefore;

2.5 moles of B will mix with (1/3)×2.5 moles of A = 5/6 moles of A

The remaining number of moles of A in the system = 2 - 5/6 =  7/6 moles of A

Similarly, we have;

At least, 0.125 mole of B combines with 1 mole of A to form a two-phase system

7/6 moles of A will combine with 7/6 × 0.125 = 7/48 moles of B to form a two-phase system

The number of moles of B left = 0.5 - 7/48 = 17/48 = 0.3541[tex]\overline 6[/tex] moles of A

Therefore, we have;

5/6 moles of A and 2.5 moles of B in the one-phase system

7/6 moles of A and 7/48 moles of B in the two-phase system

17/48 moles of B remaining in the system

Answer:

mass 1.25 Liters NH₃(gas) = 0.949 grams (3 sig-figs)

Explanation:

At STP (Standard Temperature-Pressure conditions => 0°C(=273K) and 1atm pressure,  1 mole any gas will occupy 22.4 Liters.

So, given 1.25 Liters ammonia gas at STP, convert to moles then multiply by formula wt. (17g/mole gives mass of NH₃.

moles NH₃(gas) = 1.25L NH₃(gas)/22.4L NH₄(gas)· NH₃(gas)mole⁻¹ = 0.0558 mole NH₃(gas).  

Converting to grams NH₃(gas) = 0.0558 mole NH₃(gas) x 17 g·mol⁻¹ = 0.949 grams NH₃(gas).

Answer:

m = 29.6 grams

Explanation:

Given that,

Number of moles = 0.540

The molar mass of manganese = 54.93 g/mol

We know that,

Number of moles = given mass/molar mass

[tex]m=n\times M\\\\m=0.540 \times 54.93\\\\m=29.6\ g[/tex]

So, the required mass of the Manganese is equal to 29.6 grams.

Answer:

rocky surface covered with landforms

Explanation:

Answer: You forgot to zero the balance

Explanation:

Answer:

What is their question answers exactly? I won't be able to explain who is correct.

Explanation:

Answer:

Not a lot to draw from this question if im going to be honest

Explanation