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3 years ago

________ function by absorbing excess hydrogen or hydroxide ions.

Chemistry

2 answers:

MrMuchimi3 years ago

4 0

Answer: Buffer solution

Explanation:

Buffer is a solution which resists changes in pH when small quantities of acid or alkali are added to it.

It is formed by combination of weak acid and salt of weak acid with strong base called as acidic buffer.Example: $CH_3COOH$ and $CH_3COONa$

Or it is formed by combination of weak base and salt of weak base with strong acid called as basic buffer buffer. Example: $NH_4OH$ and $NH_4Cl$

When acid is added to buffer, the buffer releases $OH^-$ ions to neutralize the acid added and when base is added to buffer, the buffer releases $H^+$ ions to neutralize the base added.

Kay [80]3 years ago

3 0

Buffers function by absorbing excess hydrogen or hydroxide ions.

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What molecules must the plant take in so the process of photosynthesis can occur

lilavasa [31]

Answer:

Glucose molecules

Explanation:

7 0

3 years ago

A bulldozer does 15,000 J of work lifting dirt 15 m up to get it out of a hole. How much force did the bulldozer use to accompli

nika2105 [10]

Answer:

<h3>The answer is 3000 N</h3>

Explanation:

The amount of force can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{15000}{5} \\$

We have the final answer as

Hope this helps you

7 0

2 years ago

DerKrebs [107]

Answer:

F = 30 N

Explanation:

Given data:

Mass of toy train = 1.5 kg

Acceleration of train = 20 m/s²

Amount of force acting on it = ?

Solution:

The net force on object is equal to the its mass multiply by its acceleration.

Formula:

F = ma

F = force

m = mass

a = acceleration

Now we will put the values in formula.

F = 1.5 kg × 20ms⁻²

F = 30 kg.ms⁻²

kg.ms⁻² = N

F = 30 N

7 0

3 years ago

what is the percent by mass of a solution that contains 30 grams of potassium nitrite in 0.5 kilograms of water?

Makovka662 [10]

Answer:

5.66 %.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

mass of potassium nitrite = 30.0 g,

mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.

∴ mass % = (mass of solute/mass of solution) x 100 = (30.0 g/530.0 g) x 100 = 5.66 %.

5 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago