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Sloan [31]
2 years ago
14

COULD SOMEONE PLEASE GIVE ME THE ANSWER FOR THIS NO BOTS PLEASEEE TODAY IS THE LAST DAY FOR MISSING WORK!!!!

Chemistry
2 answers:
Ilya [14]2 years ago
5 0

Answer:

8. D

9. A

10. D

Vikentia [17]2 years ago
3 0

Answer:

:D

Explanation:

8. D. All of the above

9. C. Mechanical weathering

10. D. All of the above

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The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

4 0
3 years ago
Which expression is equal to the number of grams (g) in 2.43 kilograms (kg)?
Gre4nikov [31]
2.43 kilograms is equal to 2430 grams because 1 kilogram is equal to 1000 grams.
4 0
3 years ago
Read 2 more answers
Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad
S_A_V [24]
Given:

Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M

Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution

Assume: 50:50 ratio of fluoride ions and HF

12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-

5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles

Assume: Volume additive

Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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3 0
3 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
How is 0.00036 written in scientific notation?
natta225 [31]
3.6 x 10^-4
I am writing this sentence because the answer does not reach the 20 character limit. Thank you have nice day! :D Cx :P
4 0
3 years ago
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