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Oliga [24]
3 years ago
8

1. The number 0.004207, when rounded off to 3 significant figures, would appear as

Chemistry
1 answer:
tresset_1 [31]3 years ago
3 0

1)

Answer:

0.00421

Explanation:

Significant figures:

The given measurement have four significant figures 1234.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

In given value the 0.00 are not significant.

when we round to three significant figures we must include 4207. The number after the third significant figure which 7 is greater than 5 so we will increase the third significant figure by one for example 0 become 1.

If the number which is next to rounded figure is 5 or grater than five the rounded figure will increase by 1.

2) In which of following  pairs of numbers does each member of the pair contain the same number of significant figures?

a) 0.0806 and 960,000

b) 71.0 and 81.00

c) 20.0 and 462

d) 0.03070 and 0.0670

Answer:

c)  20.0 and 462

Explanation:

The option c is correct because both value have same number of significant figure which are 3.

All other are incorrect because,

a) 0.0806 and 960,000

0.0806 have three significant figures and 960,000 have six significant figures .

b) 71.0 and 81.00

71.0 have three significant figures  and 81.00 have four significant figures .

d) 0.03070 and 0.0670

0.03070 have four significant figures and 0.0670 have three significant figures .

3) The mathematical meaning associated with the metric system prefixes milli , micro and nano is,

a) 10⁻², 10⁻⁴ and  10⁻⁶

b) 10⁻², 10⁻³ and 10⁻⁶

c) 10⁻³, 10⁻⁹ and 10⁻¹²

d) 10⁻³, 10⁻⁶ and 10⁻⁹

Answer:

d) 10⁻³, 10⁻⁶ and 10⁻⁹.

Explanation:

The correct option is d.

The mili is represented by "m". In metric system it is unit prefix represent thousand as 10⁻³.

This unit is adopted in 1795.

The  micro is Greek word. It is represented by symbol μ .

It is equal to 10⁻⁶.

It is presented in 1960.

The nano is equal to  10⁻⁹. It is represented by symbol "n" .

one nanometer is equal to  10⁻⁹ meter.

All other options are incorrect.

4) In which of the following measured numbers are all of zero significant?

a) 0.00480

b)0.300120

c) 6.080110

d) 99400

Answer:

option c is correct.

Explanation:

a) 0.00480

There are three significant figure in given value 480. The three zeros are not significant.

b)0.300120

There are six significant figure in given value 300120. The zero before decimal point is not significant.

c) 6.080110

There are seven significant figure in given value 6,0,8,0,1,1,0. The all zeros are significant.

d) 99400

There are three significant figure in given value 9,9,4.

5)When a substance undergoes a physical change it is always true that

a) it changes color

b) it changes state

c heat is absorbed

d) its chemical composition changes

Answer:

b) it changes state

Explanation:

Physical Change                                                                                              

The changes that occur only due to change in shape or physical properties but their chemical or internal composition remain unchanged.  

These changes were reversible.

They have same chemical property.

These changes can be observed with naked eye.

Example :

Water converting to Ice

Water converting to gas

In this water molecule remain the same only they rearrange themselves that change its state of mater not composition

6)  Object A weighs 40 grams and has a volume of 5 mL. Object B weighs 12 grams and has a volume.

Therefore,

a) A and B have equal densities.

b) B is twice as dense as A.

c) B is four times as dense as A.

d) A is more dense than B.

Answer:

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

For object A:

Mass = 40 g

Volume = 5 mL

Density = ?

Solution:

d = m/ v

d = 40 g/ 5 mL

d = 8 g/mL

The volume of object b is missing we can not determine density by just mass value.

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Fed [463]

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

7 0
3 years ago
Cecil writes the equation for the reaction of hydrogen and oxygen below.
Wittaler [7]

Answer:

C) to show that atoms are conserved in chemical reactions

Explanation:

When writing a chemical reaction, we should always consider the Mass Conservation Law, which basically states that; in an isolated system; the total mass should remain constant, this is, the total mass of the reactives should be equal to the total mass of the products

For this case, we should add the apporpiate coefficients in order to be in compliance with this law:

2H₂ + O₂ → 2H₂O

So, we can check the above statement:

For reactives (left side):

4H

2O

For product (right side):

4H

2O

5 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0
3 years ago
4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Mariulka [41]

Answer:

0.9 moles of water

Explanation:

Use mole ratios:

5 : 6

divide by 5 on both sides

1 : 1.2

multiply by 0.75 on both sides

0.75 : 0.9

So the result is 0.9 moles of water

(Please correct me if I'm wrong)

5 0
2 years ago
Chemistry question please help!!
LiRa [457]
Answer: It’s the first one
8 0
3 years ago
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