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3 years ago

WILL MARK BRAINLIEST IF YOUR ANSWER IS CORRECT

Chemistry

2 answers:

olga2289 [7]3 years ago

6 0

Answer:

1. Element

2. Compound

3. Heterogeneous mixture

4. Homogeneous mixture

5. Homogeneous mixture

6. Heterogeneous and homogeneous mixture

7. Homogeneous mixture

8. Heterogeneous mixture

Anettt [7]3 years ago

3 0

1. Gold is element
4. D
2 b
3c

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4. Explain error and how it can impact your results from an experiment

Jet001 [13]

Random errors will shift each measurement from its true value by a random amount and in a random direction. These will affect reliability (since they're random) but may not affect the overall accuracy of a result.

3 0

2 years ago

As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is

chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

KBrO₃ → K⁺ + BrO₃⁻ 

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = 38,0 kJ/mol

I hope it helps!

3 0

2 years ago

36g of an alloy of copper and zinc contains 45% copper. how much pure copper do you need to add in order to get a 60% copper all

iogann1982 [59]

First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper

7 0

3 years ago

PLEASE ANSWER THIS QUESTION ASAP.

Natali5045456 [20]

Equal is the answer

Hope this helps

6 0

2 years ago

Read 2 more answers

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

3 years ago

Read 2 more answers