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astraxan [27]
3 years ago
5

In this experiment, you will measure all masses in

Chemistry
1 answer:
Xelga [282]3 years ago
4 0

Answer:

0.498 kg

Explanation:

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How would I find the answer?
ELEN [110]

Answer:

Kr

Explanation:

The noble gas that is isoelectronic with Br⁻ is krypton.

This is because krypton is the closest noble gas to Br on the periodic table.

Electronic configuration of Bromine is;

            2, 8, 18, 7

  Br⁻ becomes;  2, 8, 18, 8

  Krypton is;       2, 8, 18, 8

4 0
3 years ago
What tool is used in an explosion site to chemically detect explosive residues on objects?
qaws [65]

Answer:

ion mobility spectrometer

Explanation:

it is an handheld chemical dectection device

Hope this helps

Please mark me as Brainliest

8 0
3 years ago
How are the environments of a desert and a tundra different?
klasskru [66]
B. A Tundra is less humid than a desert.

Reasoning: Tundra’s experience low temperatures while deserts experience hot/humid temperatures.
6 0
2 years ago
To what Celsius temperature must 67.0 mL of krypton gas at 18.0°C be changed so the volume will triple? Assume the pressure and
Mice21 [21]

Answer: 600°C

Explanation:

This reaction is explained by Charles' law as the pressure is constant.

From the question, we obtained:

V1 = 67mL

T1 = 18°C = 18 +273 = 291K

V2 = 3V1 ( Vol is tripled) = 3x67 = 201mL

T2 =?

Applying the Charles' law,

V1 /T1 = V2 /T2

67/291 = 201 / T2

Cross multiply to express in linear form.

67xT2 = 291x201

Divide both side by 67, we have:

T2 = (291x201) /67

T2 = 873K

Converting to Celsius temperature, we have

T°C = K — 273

T°C = 873 — 273 = 600°C

8 0
3 years ago
How many joules (J) are needed to increase the temperature of 10.0 g of lead (Pb) from 30 ˚C to 50 ˚C? (Cp for Pb = 0.128 J/g˚C)
MA_775_DIABLO [31]

Answer:

Q = 25.6 j

Explanation:

Given data:

Energy needed= ?

Mass of lead = 10.0 g

Initial temperature = 30 °C

Final temperature = 50°C

Cp = 0.128 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 50°C - 30°C

ΔT = 20°C

Now we will put the values in formula.

Q = 10 g × 0.128 j/g.°C × 20°C

Q = 25.6 j

7 0
3 years ago
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