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Anettt [7]
1 year ago
9

Match each tool on the left with its most likely use on the right by choosing the correct number. ruler scale telescope thermome

ter microscope 1. Measuring the weight of an apple 2. Studying a distant object 3. Studying a very small object 4. Measuring the length of a book 5. Measuring the temperature of a cup of tea
Chemistry
1 answer:
RideAnS [48]1 year ago
7 0

Answer:

1. Measuring the weight of an apple- scale.

2. Studying a distant object- telescope.

3. Studying a very small object- microscope.

4. Measuring the length of a book- ruler.

5. Measuring the temperature of a cup of tea- thermometer.

Explanation:

1. A scale or measuring scale is used to measure or take the weight of any object. It can be either an analog or digital scale as long as the measurements can be taken from it. So, <u>measuring the weight of an apple can be done on a measuring scale</u>.

2. A telescope is an instrument that is used in scientific, especially in space explorations, to study a distant object in the universe/ sky. It gives a clear image and enables a person to view things closer than they are. So, <u>a telescope can be used to study distant objects</u>.

3. <u>A microscope is used to study very different small objects</u>, such as microbes and other organisms. It enlarges the objects to be viewed and thus, enables them to be studied in detail.

4. <u>A ruler may be used to measure the length of a book</u>. It has indications or marks by which the lengths can be taken.

5. <u>A thermometer is an instrument that is used to measure the temperature of anything</u>, be it a person or even other things such as food and even a cup of tea.  

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If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072
4 0
2 years ago
Suppose you heat an oven to 400°F (about 200°C) and boil a pot of water. Which of the following explains why you would be burned
goblinko [34]

Answer: The correct answer is: the water can transfer heat to your arm more quickly than the air.

Explanation: The heat is transferred from the air or water to your arm through convection. The convective heat transfer coefficient of water is higher than the air's, so, even though the temperature of boiling water is lower, the heat will be transferred more efficiently to the other surface, in this case, the hand.

6 0
2 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
2 years ago
The adult blue whale has a lung capacity of 5.0×103 L5.0×103 L. Calculate the mass of air (assume an average molar mass 28.98 g/
andrew11 [14]

Answer:

The mass of the air is 6920.71g

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Volume (V) = 5.0x10^3 L

Molar Mass of air (M) = 28.98 g/mol

Temperature (T) = 0.2°C

Pressure (P) = 1.07 atm

mass air (m) =?

Number of mole (n) =?

Recall:

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Conversion of celsius temperature to Kelvin temperature.

K = °C + 273

°C = 0.2°C

K = °C + 273

K = 0.2°C + 273

K = 273.2 K

Therefore, the temperature (T) = 273.2 K

Step 3:

Determination of the number of mole of air.

Applying the ideal gas equation PV = nRT, the number of mole n, can be obtained as follow:

PV = nRT

1.07 x 5.0x10^3 = n x 0.082 x 273.2

Divide both side by 0.082 x 273.2

n = (1.07 x 5.0x10^3)/(0.082 x 273.2)

n = 238.81 moles

Step 4:

Determination of the mass of air. This is illustrated below:

Number of mole of air = 238.81 moles

Molar Mass of air = 28.98 g/mol

Mass of air =.?

Mass = number of mole x molar Mass

Mass of air = 238.81 x 28.98

Mass of air = 6920.71g

3 0
2 years ago
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