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lukranit [14]
3 years ago
12

1. फल

Chemistry
1 answer:
Allisa [31]3 years ago
5 0

Answer:

बालकेभ्य।

2 गुरुभि :

3 •••••

4 कवीन।

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Some animals hunt by _______ or sneaking up on their prey. It a nine letter word and there a m in the second place
guajiro [1.7K]

Answer:

Some animals hunt by <em>Camouflage</em> or sneaking up on their prey.

5 0
2 years ago
Problem: What is the kinetic energy of a 0.500 kg rabbit, running at 8.00 m/s? Calculate using the formula,
arsen [322]

Answer:

the answer s 9.000ms

Explanation:

that's the answer

4 0
2 years ago
What energy keeps the water cycle going?
Elden [556K]

Answer:

the the that keeps the water cycle going is heat

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3 years ago
Read 2 more answers
A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside
Inessa05 [86]
<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

  • Initial volume, V1 = 100.0 mL
  • Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

  • Initial pressure, P1 = 1.80 atm
  • Final temperature , T2 = -25°C

                                     = 248.15 K

  • Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

  • According to the combined gas law equation;

\frac{P1V1}{T1}=\frac{P2V2}{T2}

Rearranging the formula;

V2=\frac{P1V1T2}{T1P2}

Therefore;

V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}

V2=112.08mL

Therefore, the new volume of the gas is 112.08 mL

8 0
3 years ago
500ml of a buffer solution contains 0.050 mol nahso3 and 0.031
nydimaria [60]

Answer:

The answers are explained below

Explanation:

a)

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

b)

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

c)

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

d)

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

e)

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

4 0
3 years ago
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