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3 years ago

What can you conclude about why Arrhenius bases are considered a limited case of Bronsted-Lowry bases.

Physics

2 answers:

VLD [36.1K]3 years ago

8 0

Answer:

the 3rd one

Explanation:

iragen [17]3 years ago

8 0

Answer: C
Hope this help!!!

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Air friction damps a tuning fork so the amplitude decreases by 1/10 per second. Resonances] 10 Hz. what % chnge in frequency res

aleksandr82 [10.1K]

Answer:

The percentage change in frequency is 10%.

Explanation:

Given that,

Amplitude decreases by 1/10 per second.

Resonance = 10 Hz

We need to calculate the change in frequency

Using formula of change in frequency

$\Delta f=\Delta A\times R$

Put the value into the formula

$\Delta f=\dfrac{1}{10}\times10$

$\Delta f=1\ Hz$

We need to calculate the resonant frequency

Using formula of resonant frequency

$f_{r}=R-\Delta f$

Put the value into the formula

$f_{r}=10-1$

$f_{r}=9\ Hz$

We need to calculate the percentage change in frequency

Using formula of percentage change in frequency

$f=\dfrac{10-9}{10}$

$f=10\%$

Hence, The percentage change in frequency is 10%.

7 0

4 years ago

A car could move at constant speed on an icy curve which is banked for _______________ (all, one, no) speed(s) of the car.

son4ous [18]

All is the answer I believe.

4 0

3 years ago

Helpp for number 6 please

Vanyuwa [196]

A neutral object contains no net charge, but still has many charged particles within the object, it just means that the charges cancel each other out.

7 0

3 years ago

After your patched the roof, you dropped the hammer off the house and you heard it land on your toolbox on the ground 3.3s later

Schach [20]

Answer:

53.36 m.

Explanation:

Initial velocity of the toolbox u = 0

acceleration due to gravity g = 9.8 m s⁻²

time t = 3.3 s

height of roof h = ?

h = ut + 1/2 g t²

= 0 + .5 x 9.8 x 3.3²

= 53.36 m.

3 0

4 years ago

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

3 0

4 years ago