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kolezko [41]
1 year ago
13

As part of an investigation, students combined substances in a beaker to observe chemical reactions. They performed two procedur

es. They measured the mass of each substance before and after each reaction. The table shows their observations.
Assuming the students did not make any careless errors, what likely explains these changes in mass?

A. Procedure 1: All the reactants were liquids that evaporated.
Procedure 2: A gas was formed as one product, and it escaped into the air.

B. Procedure 1: One of the reactants was converted to thermal energy.
Procedure 2: All the products were liquids.

C. Procedure 1: The reactants were liquids with different densities.
Procedure 2: The reactants were combined into only one product.

D. Procedure 1: One of the products was a gas that escaped into the air.
Procedure 2: A gas from the air reacted with one of the other reactants.

Chemistry
2 answers:
asambeis [7]1 year ago
8 0

Answer:

D.

Explanation:

aniked [119]1 year ago
3 0

Answer: its D

Explanation: I took a test, got it correct.

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Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of
Andrej [43]

Given the mass of hydrogen = 16.0 oz

Converting 16.0 oz hydrogen to pounds (lb) using the conversion factor 1 lb = 16 oz:

16.0 oz * \frac{1 lb}{16 oz} =1 lb

Converting 16.0 lb to g using the conversion factors 1 kg = 2.2 lb, 1 kg = 1000 g:

1lb * \frac{1kg}{2.2lb}*\frac{1000g}{1kg}= 454.5 g

Heat of combustion of hydrogen = 142 J/g

Calculating the heat released when 16.0 oz is combusted:

454.5g H_{2} * \frac{142 J}{g} *\frac{1 kJ}{1000J}=64.5kJ


5 0
1 year ago
Read 2 more answers
What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?
kirza4 [7]
We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution 

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

                         = 3.5 g / 58.44 

                         = 0.0599 moles


when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg 

now we can get the molality = moles NaCl / Kg water

                                                =0.0599moles/0.23Kg

                                                = 0.26 m

∴T2-0 =  - 2 * 0.26 *1.86

∴T2 = -0.967 °C
8 0
2 years ago
Read 2 more answers
Sulfurous acid, h2so3, breaks down into water (h2o) and sulfur dioxide (so2). if only one molecule of sulfurous acid was involve
devlian [24]

Explanation:

The given reaction equation will be as follows.

  H_{2}SO_{3} \rightarrow H_{2}O + SO_{2}

Now, number of atoms on reactant side are as follows.

  • H = 2
  • S = 1
  • O = 3

Number of atoms on product side are as follows.

  • H = 2
  • S = 1
  • O = 3

Therefore, this equation is balanced since atoms on both reactant and product sides are equal.

Thus, we can conclude that there is one sulfur atom in the products.

6 0
2 years ago
Read 2 more answers
Predict what will observe in below mention experiment.
Sloan [31]

Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature

a. It is likely that more rock candy will be formed in batch A.

b. It is likely that less rock candy will be formed in batch A.

c. It is likely that no rock candy will be formed in either batch.

d. I need more information to predict which batch is more likely to form rock candy.

Answer: Option A

Explanation:

More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.

Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.

The sugar will be dissolved in water until the time all the space is filled sugar molecules.

Hence, the correct answer is Option A.

6 0
2 years ago
Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
1 year ago
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