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Alenkasestr [34]
3 years ago
13

7. Halfway through a cross country meet, a

Chemistry
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

3m/s

Explanation:

Data obtained from the question include:

Initial speed (s1) = 4 m/s

Final speed (S2) = 7m/s

Change in speed (ΔS)

ΔS = s2 — s1

ΔS = 7 — 4

ΔS = 3m/s

Therefore, the change in speed is 3m/s

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Вычислить равновесные концентрации частиц в растворе, содержащем 0,01 моль/л Cu(NO3)2 и 1,0 моль/л NH3.
Wewaii [24]

Answer:

i dOnT SpEaK uR lAnGuAgE

Explanation:

ReEEEeEeEeEeEeEeEeEeeEEEeeeeEEEEeeeeeEEeEEeeeeeEEEEEeeeEee

3 0
3 years ago
When all colors are absorbed, you see:<br><br> black<br> white<br> green<br> yellow
wolverine [178]

uhmm, white.

Explanation:

you'll basically look like blind ig

5 0
3 years ago
How much heat will be released when 10.0 g of hydrogen peroxide decomposes according to the following reaction: 2H2O2(l) 2H2O(l)
kramer
C. 28 KJ

AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2

0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
5 0
3 years ago
Provided is a diagram showing forces on a box.
Eduardwww [97]

Answer:

5 N left

Explanation:

trust me cccuuhh

4 0
3 years ago
If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
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