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3 years ago

Calculate the average time it took the car to travel 0.25 What is the average time it took the car to travel 0.25

Chemistry

2 answers:

DedPeter [7]3 years ago

8 1

Answer:

Explanation:

idk why all u guys like trump who do.. hes just a big pain in the a*s.

blaze the wolf

2 years ago

shut the f*ck up this is for work not your bullsh*t

blaze the wolf

2 years ago

biden is a f*cking dumb b*tch im only a child and i can run a country better than his old slow 2 mph walking a*s

yawa3891 [41]3 years ago

4 0

Answer:

2.23 3.13

Explanation:

just did it

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What is the Net Ionic equation for this chemical reaction: FeBr2+Na2S=FeS+2NaBr

Artyom0805 [142]

Answer: Fe(aq)+S(aq)=FeS(s)

Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:

1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)

Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:

Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr

2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.

3.) Ionic formula: Fe(aq)+ 2Br(aq)+2 Na(aq)+S (aq)=FeS(s)+2 Na(aq)+2Br(aq)

4.)When you've derived a total ionic equation (above), you'll find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)

Hope this helps :)

7 0

3 years ago

The osmotic pressure of a saturated solution of strontium sulfate at 25 ∘C∘C is 21 torrtorr. Part A What is the solubility produ

USPshnik [31]

Answer

solubility product = 3.18x 10^-7

Explanation:

We were given the pressure in torr then we need to convert to atm for consistency, ten we have

21torr/760= 0.0276315789 atm

21 Torr = .0276315789 atm

P = i M S T

M = P / iRT

Where p is osmotic pressure

T= temperature= 25C+ 273= 298K

for XY vanthoff factor i = 2

S = 0.0821 L-atm / mol K

M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)

M = 0.000564698046 mol/liters

solubility= 0.000564698046 mol/liters

Ksp = [X+][Y-]

Ksp = X^2

Ksp = [Sr^+2] * [SO4^-2]

Ksp = X^2

Ksp = (0.000564698046)^2

Ksp = 3.18883883 × 10-7

Ksp = 3.18x 10^-7

solubility product = 3.18x 10^-7

Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7

7 0

3 years ago

Write the electron configuration for the following elements:

vazorg [7]

Answer:

a.Carbon [He]2s22p2

b. Neon [He]2s22p6

c. Sulfur [Ne]3s23p4

d.Lithium [He]2s1

e. Argon [Ne]3s23p6

f. Oxygen [He]2s22p4

g.Potassium [Ar]4s1

h. Helium 1s2

This table is available to download as a PDF to use as a study sheet.

NUMBER ELEMENT ELECTRON CONFIGURATION

1 Hydrogen 1s1

2 Helium 1s2

3 Lithium [He]2s1

4 Beryllium [He]2s2

5 Boron [He]2s22p1

6 Carbon [He]2s22p2

7 Nitrogen [He]2s22p3

8 Oxygen [He]2s22p4

9 Fluorine [He]2s22p5

10 Neon [He]2s22p6

11 Sodium [Ne]3s1

12 Magnesium [Ne]3s2

13 Aluminum [Ne]3s23p1

14 Silicon [Ne]3s23p2

15 Phosphorus [Ne]3s23p3

16 Sulfur [Ne]3s23p4

17 Chlorine [Ne]3s23p5

18 Argon [Ne]3s23p6

19 Potassium [Ar]4s1

20 Calcium [Ar]4s2

21 Scandium [Ar]3d14s2

22 Titanium [Ar]3d24s2

23 Vanadium [Ar]3d34s2

24 Chromium [Ar]3d54s1

25 Manganese [Ar]3d54s2

26 Iron [Ar]3d64s2

27 Cobalt [Ar]3d74s2

28 Nickel [Ar]3d84s2

29 Copper [Ar]3d104s1

30 Zinc [Ar]3d104s2

31 Gallium [Ar]3d104s24p1

32 Germanium [Ar]3d104s24p2

33 Arsenic [Ar]3d104s24p3

34 Selenium [Ar]3d104s24p4

35 Bromine [Ar]3d104s24p5

36 Krypton [Ar]3d104s24p6

37 Rubidium [Kr]5s1

38 Strontium [Kr]5s2

39 Yttrium [Kr]4d15s2

40 Zirconium [Kr]4d25s2

41 Niobium [Kr]4d45s1

42 Molybdenum [Kr]4d55s1

43 Technetium [Kr]4d55s2

44 Ruthenium [Kr]4d75s1

45 Rhodium [Kr]4d85s1

46 Palladium [Kr]4d10

47 Silver [Kr]4d105s1

48 Cadmium [Kr]4d105s2

49 Indium [Kr]4d105s25p1

50 Tin [Kr]4d105s25p2

51 Antimony [Kr]4d105s25p3

52 Tellurium [Kr]4d105s25p4

53 Iodine [Kr]4d105s25p5

54 Xenon [Kr]4d105s25p6

55 Cesium [Xe]6s1

56 Barium [Xe]6s2

57 Lanthanum [Xe]5d16s2

58 Cerium [Xe]4f15d16s2

59 Praseodymium [Xe]4f36s2

60 Neodymium [Xe]4f46s2

61 Promethium [Xe]4f56s2

62 Samarium [Xe]4f66s2

63 Europium [Xe]4f76s2

64 Gadolinium [Xe]4f75d16s2

65 Terbium [Xe]4f96s2

66 Dysprosium [Xe]4f106s2

67 Holmium [Xe]4f116s2

68 Erbium [Xe]4f126s2

69 Thulium [Xe]4f136s2

70 Ytterbium [Xe]4f146s2

71 Lutetium [Xe]4f145d16s2

72 Hafnium [Xe]4f145d26s2

73 Tantalum [Xe]4f145d36s2

74 Tungsten [Xe]4f145d46s2

75 Rhenium [Xe]4f145d56s2

76 Osmium [Xe]4f145d66s2

77 Iridium [Xe]4f145d76s2

78 Platinum [Xe]4f145d96s1

79 Gold [Xe]4f145d106s1

80 Mercury [Xe]4f145d106s2

81 Thallium [Xe]4f145d106s26p1

82 Lead [Xe]4f145d106s26p2

83 Bismuth [Xe]4f145d106s26p3

84 Polonium [Xe]4f145d106s26p4

85 Astatine [Xe]4f145d106s26p5

86 Radon [Xe]4f145d106s26p6

87 Francium [Rn]7s1

88 Radium [Rn]7s2

89 Actinium [Rn]6d17s2

90 Thorium [Rn]6d27s2

91 Protactinium [Rn]5f26d17s2

92 Uranium [Rn]5f36d17s2

93 Neptunium [Rn]5f46d17s2

94 Plutonium [Rn]5f67s2

95 Americium [Rn]5f77s2

96 Curium [Rn]5f76d17s2

97 Berkelium [Rn]5f97s2

98 Californium [Rn]5f107s2

99 Einsteinium [Rn]5f117s2

100 Fermium [Rn]5f127s2

101 Mendelevium [Rn]5f137s2

102 Nobelium [Rn]5f147s2

103 Lawrencium [Rn]5f147s27p1

104 Rutherfordium [Rn]5f146d27s2

105 Dubnium *[Rn]5f146d37s2

106 Seaborgium *[Rn]5f146d47s2

107 Bohrium *[Rn]5f146d57s2

108 Hassium *[Rn]5f146d67s2

109 Meitnerium *[Rn]5f146d77s2

110 Darmstadtium *[Rn]5f146d97s1

111 Roentgenium *[Rn]5f146d107s1

112 Copernium *[Rn]5f146d107s2

113 Nihonium *[Rn]5f146d107s27p1

114 Flerovium *[Rn]5f146d107s27p2

115 Moscovium *[Rn]5f146d107s27p3

116 Livermorium *[Rn]5f146d107s27p4

117 Tennessine *[Rn]5f146d107s27p5

118 Oganesson *[Rn]5f146d107s27p6

Explanation:

I hope it's help

8 0

2 years ago

How is the pH of an acidic substance raised?

BabaBlast [244]

Answer: 4) Concentrate it

Explanation:

4 0

2 years ago

When the following equation is balanced, the coefficients are ________.

LuckyWell [14K]

Answer:

Explanation:

Option B is the correct answer. When you add 7 in front of the O2 in the left side, and then you add 4 and 6 in front of NO2 and H2O respectively, the number of oxygen atoms is the same in both sides.

4 0

2 years ago