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AnnZ [28]
3 years ago
14

HELP!!! WILL RECEIVE BRAINLIEST!!!!! What is the independent variable

Chemistry
2 answers:
Vsevolod [243]3 years ago
5 0
An independent variable Is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable, In that case - 1. sulfate solution, 2. potassium iodide solution, 3. hydrochloric acid solution.
ycow [4]3 years ago
3 0
You kind of need to show more, but the independent variable would be something that you are changing to then measure the reaction, like the time left in for example.
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A solid keeps its shape due to which of the following factors? (3 points)
algol [13]

Answer: Attractive forces between particels

Explanation:

8 0
3 years ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0
3 years ago
A model of an ideal gas is shown here. A student designs an experiment to test the impact of increasing the pressure on the syst
k0ka [10]

A. As pressure on the gas increases, the volume and temperature will both decrease

6 0
3 years ago
What characteristics do the elements in a group share? In a period, share?
Katyanochek1 [597]

Answer:

The elements in same period have same principle quantum number or energy shell.

The elements in same group shows similar chemical and properties.

Explanation:

Inn group:

The elements in same group i.e present in vertical column shows similar chemical properties.

The elements in same group having same number of valance electrons. while in chemical reaction bonds are break and formed and valance electrons are involved. That's why elements in same group having same number of valance electrons and shows similar chemical properties.

In period:

While as we move from left to right the number of valance electron increase by one in every element. But the electron is added in same shell which means that their physical and chemical properties are different but principal quantum number is same.

3 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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