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eduard
3 years ago
14

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCl. What volume of 3.00 M HCl did the stu

dent use to make the sample? Use mc020-1.jpg. 3.70 mL 16.7 mL 30.0 mL 83.3 mL
Chemistry
2 answers:
ivann1987 [24]3 years ago
7 0
Um in  not sure but i think is=t should be 60.2874

Yuliya22 [10]3 years ago
7 0
Use the dilution formula C1V1= C2V2

C1= 3.0 M
V1= ??
C2= 1.80 M
V2= 50.0mL

(3.0 M) x (V1)= (1.80 M) x (50.0 mL)

V1= 30.0 mL. 
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Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
HELP!!!! I WILL MARK BRAINLIEST WHOEVER ANSWERS FIRST
ELEN [110]

Answer:

I answered first, give me brainliest

Explanation:

8 0
3 years ago
28Which is true?
sergij07 [2.7K]
28) All elements are made of atoms

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8 0
3 years ago
A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I
murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE  DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0
3 years ago
How does heat, without causing melting, damage a rock below eath’s surface
zheka24 [161]
It could be erosion
4 0
3 years ago
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