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Luba_88 [7]
3 years ago
6

Please help with this giving brainlist !!

Chemistry
1 answer:
mash [69]3 years ago
6 0

Answer:

if im right i do my caculation that it is B

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• 1) The pressure of a sample of gas in a 2.00-L container is 876 mmHg.
kifflom [539]

Answer:

4380 mmHg

Explanation:

Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:

P1V1=P2V2

In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:

P1V1= P2V2

876 mmHg * 2L = P2 * 400ml /(1000ml/L)

P2= 876 mmHg * 2L / 0.4L

P2= 4380 mmHg

5 0
3 years ago
Net ionic of ammonium sulfide added to iron (ll) chloride
meriva

Answer:

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)  

Step-by-step explanation:

Molecular Equation:

(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)

Ionic equation :

2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)

Net ionic equation :

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)

4 0
3 years ago
Given a stock standard solution of concentration 0.200 M, calculate the volume in mL required to make up calibration standards o
guapka [62]

M1V1 = M2V2

.200 (.025) = 1.60 X 10 -2 (V2)

V2 = .315 L

1.60 x 10-2 M  in 315 mL

6 0
3 years ago
A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
What distinguishes physical properties from chemical properties
ale4655 [162]

A physical property is an aspect of matter that can be observed or measured without changing it. A chemical property may only be observed by changing the chemical identity of a substance.


6 0
3 years ago
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