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2 years ago

Jim thinks that he can throw a baseball six meters per second. What does the six meters refer to? (2 points)

Chemistry

2 answers:

belka [17]2 years ago

7 0

The answer is b (the distance the baseball travels in one second)

Tcecarenko [31]2 years ago

6 0

The answer is b. hope this helps

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A gas occupies 10.2 liters at 0.60 atm. What is the pressure if the volume becomes 3.0 L?

Arturiano [62]

P1V1 = P2V2
P1 = 0.60 atm
V1 = 10.2 L
P2 = ?
V2 = 3.0 L

Solve for P2 —> P2 = P1V1/V2 = (0.60 atm)(10.2 L) / (3.0 L) = 2.04 atm

3 0

3 years ago

List four metals that will mot replace hydrogen in an acid

sdas [7]

The four metals that will not replace hydrogen in an acid are Cu, Ag, Au, Hg.

Explanation:

As Copper Cu lies below the hydrogen, this is why it is less reactive and ca not replace Hydrogen from its compound.

Silver Ag also lies below the hydrogen and therefore it is less reactive to displace Hydrogen.

Gold Au becomes less reactive than Hydrogen due to its placement below Hydrogen acid.

Mercury Hg is less reactive than Hydrogen

7 0

3 years ago

Can you help me?

bagirrra123 [75]

1. brain
2. spinal cord
3. central
4. peripheral
5. nerves
6. network
7. muscles
8. organs
9. interprets
10. transmits

this was sort of a guess so im very sorry if it was wrong !!

6 0

3 years ago

Read 2 more answers

Below is a proposed mechanism for the decomposition of H2O2. H2O2 + I– → H2O + IO– slow H2O2 + IO– → H2O + O2 + I– fast Which of

Savatey [412]

Answer: Option (d) is the correct answer.

Explanation:

The given equations as as follows.

$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$ (slow)

$H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + O_{2} + I^{-}$

Therefore, overall reaction equation will be as follows.

$2H_{2}O_{2} + I^{-} + IO^{-} \rightarrow 2H_{2}O + O_{2} + IO^{-} + I^{-}$

So, cancelling the spectator ions then the equation will be as follows.

$H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$

As, it is known that slow step of a reaction is the rate determining step. Therefore, rate law for the slow step will be as follows.

$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$

Rate law = $k[H_{2}O_{2}][I^{-}]$

Hence, the reaction is first order with respect to $[I^{-}]$ and it is also first order reaction with respect to $[H_{2}O_{2}]$ .

Also, $[I^{-}]$ acts as a catalyst in the reaction.

Thus, we can conclude that the incorrect statement is $IO^{-}$ is a catalyst.

6 0

3 years ago

Calculate the solubility at 25°c of cubr in pure water and in a 0.0030m cobr2 solution. you'll find ksp data in the aleks data t

Reil [10]

Answer:

the solubility at 25°c of cubr in pure water = 0.011 g/L

The solubility at 25°c of cubr in a 0.0030m cobr2 solution = 0.00014 g/L

Explanation:

Step 1: Data given

Ksp of CuBr = 6.27 × 10^-9

Molar mass CuBr = 143.45 g/mol

Step 2: Calculate the solubility at 25°c of cubr in pure water

Ksp = [Cu+][Br-]

The initial concentration

[Cu+] = 0M

[Br-] = 0M

The concentration at the equilibrium

[Cu+] = X M

[Br-] = X M

Ksp =6.27 * 10^-9 = X * X = X²

S = 7.9 *10^-5 mol /l

7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L

Step 3: Calculate the solubility at 25°c of cubr in a 0.0030m cobr2 solution.

The balanced equation:

CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)

For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-

For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-

The initial concentration

[Cu+] = 0M

[Br-] = 0.0060 M

At the equilibrium

[Cu+] = X

[Br-] = 0.0060 + X

Ksp = 6.27 * 10^-9 = X * (0.0060+X)

6.27 *10^-9 = 0.0060 X + X²

X = 1.0 * 10^-6 mol/L

1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L

6 0

3 years ago