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Furkat [3]
3 years ago
9

A researcher wants to test the solubility (property of being dissolved) of salt in water as the temperature of the water increas

es.
Design an experiment to test his hypothesis that as the water temperature increases the solubility of the salt in water also increases. Include the following in your experimental design: experimental setup, procedure for data collection methods and criteria for evaluating the hypothesis.



Materials: salt, water, beakers, heat source, thermometer, balance
Chemistry
1 answer:
aev [14]3 years ago
6 0
Have about 5 beakers all with different temperatures of water. Put in a teaspoon of salt at a time and when it stops dissolving stop adding and record how much salt it took. It should be more salt as the temperature rises. The independent variable is the waters temperature. The dependent variable is how much salt is used. Make sure that there is the same amount of water in each beaker. Or else it won’t work.
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Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

n=\frac{m}{M_m}

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol

Now we can find the pressure of the gas by using the ideal gas law:

pV=nRT

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm

8 0
4 years ago
6.How many moles of gas would be in contained in a 11.2 L container that is at a
qaws [65]

0.34 moles of gas would be contained in a 11.2 L container that is at a pressure of 0.75 atm and 300 K.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

The number of moles of a substance can be calculated using the following expression:

PV = nRT

Where;

  • p = pressure (atm)
  • v = volume (L)
  • n = number of moles
  • R = gas law constant
  • T = temperature

0.75 × 11.2 = n × 0.0821 × 300

8.4 = 24.63n

n = 8.4 ÷ 24.63

n = 0.34 moles

Therefore, 0.34 moles of gas would be contained in a 11.2 L container that is at a pressure of 0.75 atm and 300 K.

Learn more about number of moles at: brainly.com/question/1190311

4 0
2 years ago
Crystals most often form in igneous and metamorphic rocks. According to the rock cycle, which processes and features do these ro
Zielflug [23.3K]

The features that are shared by the rocks include the melting and cooling of rocks.

<h3>What is a rock? </h3>

It should be noted that a rock simply means a relatively hard naturally occurring mineral material.

Crystals most often form in igneous and metamorphic rocks. The processes that these rocks share that likely promote the formation of crystals is the melting and cooling of rocks.

Learn more about rocks on:

brainly.com/question/398139

4 0
2 years ago
Which type of perspective makes objects look similar to how they look in real life (select the best answer)
olasank [31]
I believe it's fractional perspective
5 0
3 years ago
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
3 years ago
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