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3 years ago

After each step in a dichotomous key, the questions get ________.

Chemistry

2 answers:

erica [24]3 years ago

8 0

I think it’s the 1 one lmk if it’s correct

GenaCL600 [577]3 years ago

4 0

Answer:

I believe the answer is 1. more specific.

Explanation:

Since the dichotomous key is used to identify a species, you would have to get down to the specifics, so specificity is required.

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How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?

choli [55]

Answer:

r u in high school this is hard

Explanation:

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3 years ago

Show the difference between a saturated hydrocarbon and an unsaturated hydrocarbon.

harkovskaia [24]

Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.

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3 years ago

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

diamong [38]

Answer:

$\large \boxed{\text{98 m$\cdot$s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}$

5 0

4 years ago

calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg

alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

learn more about Ions:

brainly.com/question/13692734

#SPJ4

6 0

2 years ago

What is the condition of the outside air at a certain time and place?

11111nata11111 [884]

Weather... weather is the obvious answer

6 0

4 years ago

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