Answer:
Mintu should chosen a material with a lower specific heat
Explanation:
I got the answer right on the test trust me it is correct
7. A cyclist starting from rest accelerates at a rate of 6 m/s until she reaches 30 m/s. How long
does it take her to accelerate?
Answer:
Explanation:
The given values are:
a= 6m/s²
u= 0 m/s
v = 30 m/s
t= ?
The formula is :
[tex]a=\frac{v-u}{t}[/tex]
thus,
[tex]t=\frac{v-u}{a}\\t=\frac{30-0}{6}\\t= 5 seconds[/tex]
Which answer below correctly identifies the type of change and the explination for the boiling of water
which is not possible sign of chemical change
A:color change
B:light given off
C:heat given off
D:a change of state
Answer:
A. color change
Explanation:
changing the color of a metal does not change its physical properties.
hope it helps.
Which two particles are found in an atom's nucleus?
O A. Neutrons
O B. Electrons
O C. Protons
O D. Isotopes
Answer:
protons and neutrons so A and C
The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen they can deliver. Four common nitrogen-containing fertilizers are ammonia, ammonium nitrate, ammonium sulfate, and urea [(NH2)NH2) 2CO2CO)]. Rank these fertilizers in terms of the mass percentage nitrogen they contain. Rank fertilizers from largest to smallest mass percentage nitrogen they contain. To rank fertilizers as equivalent, overlap them.
Answer:
Ammonia > Urea > Ammonium nitrate > Ammonium sulphate
Explanation:
Percentage by mass of nitrogen in NH3:
Molar mass of NH3= 17 g/mol
Hence % by mass = 14/17 × 100 = 82.35%
% by mass of NH4NO3
Molar mass of NH4NO3 = 80.043 g/mol
Hence; 28/80.043 × 100 = 34.98%
% by mass of (NH4)2SO4;
Molar mass of (NH4)2SO4= 132.14 g/mol
Hence; 28/132.14 × 100 = 21.19%
% by mass of CH4N2O
Molar mass of urea = 60.0553 g/mol
Hence 28/60.0553 × 100 = 46.62%
Danielle is asked to determine the density and the identity of a metal. She determines the mass of the metal to be 102.06g and then determines the volume of the sample using water displacement. Find the density. Round the answer to the nearest whole number?
Answer:
11 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of metal = 102.06 g
Volume of water = 38 mL
Volume of water + metal = 47 mL
Density of metal =.?
Next, we shall determine the volume of the metal. This can be obtained as shown below:
Volume of water = 38 mL
Volume of water + metal = 47 mL
Volume of metal =..?
Volume of metal = (Volume of water + metal) – (Volume of water)
Volume of metal = 47 – 38
Volume of metal = 9 mL
Finally, we shall determine the density of the metal as follow:
Mass of metal = 102.06 g
Volume of metal = 9 mL
Density of metal =.?
Density = mass /volume
Density of metal = 102.06 / 9
Density of metal = 11.34 ≈ 11 g/mL
Therefore, the density of the metal is 11 g/mL
which element would be used to help support a building
C
He
Na
Fe
What will an object moving at a constant velocity do?
come to a stop, unless it is pushed by another force
A.increase in speed
B.maintain a constant velocity until acted on by another force
C.come to a stop on its own
Answer:B
Explanation: an object in motion will stay in motion until acted on by another force and an object in rest will stay in test until acted on by another force
Choose all the answers that apply. Which of the following energy forms are involved in a nuclear power plant?
1)heat
2)mechanical
3)electrical
4)nuclear
Answer:Three mutual conversions of energy forms occur at nuclear power plants: nuclear energy is converted into thermal energy, thermal energy is converted into mechanical energy, and mechanical energy is converted into electric energy.
Explanation:
N2(g) + 2H(g) -> N2 H4(g) What are the volumes of N2 gas and H2 gas required to form 28.5 grams of N2 H4 at 30'C and 1.50 atm?
Answer:
Volume of N₂ = 14.76 L
Volume of H₂ = 29.52 L
Explanation:
Given data:
Mass of N₂H₄ formed = 28.5 g
Pressure = 1.50 atm
Temperature = 30°C (30+273 = 303 k)
Volume of N₂ and H₂ needed = ?
Solution:
Chemical equation:
N₂ + 2H₂ → N₂H₄
Number of moles of N₂H₄ formed = mass/ molar mass
Number of moles of N₂H₄ formed = 28.5 g/ 32 g/mol
Number of moles of N₂H₄ formed = 0.89 mol
Now we will compare the moles of N₂H₄ with N₂ and H₂ form balance chemical equation.
N₂H₄ : N₂
1 : 1
0.89 : 0.89
N₂H₄ : H₂
1 : 2
0.89 : 2×0.89 = 1.78 mol
Volume of H₂:
PV = nRT
1.50 atm × V = 1.78 mol × 0.0821 atm.L/mol.K × 303 K
V = 44.28atm.L /1.50 atm
V = 29.52 L
Volume of N₂:
PV = nRT
1.50 atm × V = 0.89 mol × 0.0821 atm.L/mol.K × 303 K
V = 22.14 atm.L /1.50 atm
V = 14.76 L
One of the physical properties of bases is that they ____.
Question 17 options:
taste sour
feel slippery
crumble
corrode metal
Answer:
feel slippery
Explanation:they physical properties are due to its slippery nature when touched
stions and Problems What is the freezing point of Salol?
Answer:
106.7 degrees f
Explanation:
30 points!!! please help chem question
Answer: below -45
Explanation:
What is a hypothesis? O a part of the scientific method that is not necessary a specific and testable question O the variables I observe during the experiment HELP HELP PLS PLS PLS HELP NoW NowWWwWw I'll mark u BRAINLIEST!!!!!!!!
The density of two liquids, A and B, are 1000. kg/m3 and 600. kg/m3, respectively. The two liquids are mixed in a certain proportion and the density of the resulting liquid is 850 kg/m3. How much of liquid B (in grams) does 1.0 kg of the mixture contain? Assume the volume of the two liquids is additive when mixed.
Answer:
Mass of liquid B = 271.2 gram
Explanation:
Given:
Density of liquid A = 1000 kg/m³
Density of liquid B = 600 kg/m³
Density of mixture = 850 kg/m³
Mass of mixture = 1 kg
Assume:
Volume of liquid A = Va
Volume of liquid B = Vb
So,
Volume of mixture = Va + Vb
Mass of liquid A = 1000(Va)
Mass of liquid B = 600(Vb)
Mass of mixture = Mass of liquid A + Mass of liquid B
1 = 1000(Va) + 600(Vb)
Volume of mixture = 1 / 850
So,
(1/850) = Va + Vb
Vb = (1/850) - Va
1 = 1000(Va) + 600[(1/850) - Va]
Va = 7.25 × 10⁻⁴
Vb = (1/850) - Va
Vb = (1/850) - [7.25 × 10⁻⁴]
Vb = 4.25 × 10⁻⁴
Mass of liquid B = 600(Vb)
Mass of liquid B = 600(4.25 × 10⁻⁴)
Mass of liquid B = 271.2 gram
The mass of liquid B in the 1 Kg mixture is 300 g
Let the volume of liquid A be Va
Let the volume of liquid B be Vb
From the question given above, the following data were obtained:
Density of liquid A = 1000 kg/m³
Density of liquid B = 600 kg/m³
Density of mixture = 850 kg/m³
Mass of mixture = 1 kg
Mass of Liquid B (in grams) =? Next, we shall determine the mass of liquid A and BFor Liquid A:
Density of liquid A = 1000 kg/m³
Volume of liquid A = Va
Mass of Liquid A =?
Mass = Density × Volume
Mass of Liquid A = 1000 × Va
Mass of liquid A = 1000(Va)For Liquid B:
Density of liquid B = 600 kg/m³
Volume of liquid B= Va
Mass of Liquid B =?Mass = Density × Volume
Mass of Liquid B = 600 × Va
Mass of liquid B = 600(Vb)Next, we shall determine the volume of the mixture.Density of mixture = 850 kg/m³
Mass of mixture = 1 kg
Volume of mixture =?Volume = mass /density
Volume of mixture = 1 / 850
Volume of mixture = 0.0012 m³Next, we shall determine the volume of liquid AMass of mixture = 1 Kg
Mass of mixture = Mass of liquid A + Mass of liquid B
1 = 1000(Va) + 600(Vb) ....... (1)Volume of mixture = Va + Vb
Volume of mixture = 0.0012 m³
Thus,
0.0012 = Va + Vb
Vb = 0.0012 – Va ........ (2)Substitute Vb in equation 2 into equation (1)
1 = 1000(Va) + 600[0.0012 – Va]
1 = 1000Va + 0.72 – 600Va
Collect like terms
1 – 0.72 = 1000Va – 600Va
0.28 = 400Va
Divide both side by 400
Va = 0.28 / 400
Va = 0.0007 m³Next, we shall determine the volume of the liquid BVolume of mixture = 0.0012 m³
Volume of liquid A (Va) = 0.0007 m³
Volume of liquid B (Vb) =?Vb = 0.0012 – Va
Vb = 0.0012 – 0.0007
Vb = 0.0005 m³Finally, we shall determine the mass of liquid BVolume of liquid B (Vb) = 0.0005 m³
Mass of liquid B = 600(Vb)
Mass of liquid B = 600(0.0005)
Mass of liquid B = 0.3 Kg
Multiply by 1000 to express in grams
Mass of liquid B = 0.3 × 1000
Mass of liquid B = 300 gLearn more: https://brainly.com/question/17769713
Which unit of size are the smallest organisms found on Earth?
millimeters
picometers
nanometers
micrometers
Answer:
it is nanometers the third one
Explanation:
hope this helps
The correct answer is:
(D) Micrometers
Hope this helps!
The following initial rate data are for the reaction of nitrogen dioxide with carbon monoxide:
NO2 + CO --> NO + CO2
Experiment [NO2]o, M [CO]o, M Initial Rate, Ms-1
1 0.374 0.300 8.39×10-2
2 0.374 0.600 8.39×10-2
3 0.749 0.300 0.337
4 0.749 0.600 0.337
Complete the rate law for this reaction in the box below.
Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n
Rate= ________________
From these data, the rate constant is _________ M-1s-1.
Answer:
7.47 × 10^-1 M-1 s-1
Explanation:
Using the form k[A]^m[B]^n we arrive at the fact that the rate constant of the reaction can be written in the form;
Rate = k[NO2] [CO] where both m and n =1
We can use data from any of the experiments to establish the rate constant hence;
From experiment 1
[NO2] = 0.374M
[CO] = 0.300 M
R= 8.39 × 10^-2
Hence k =8.39 × 10^-2 / [0.374] [0.300]
k= 7.47 × 10^-1 M-1 s-1
*
2. Which of the following is an example of mass?
O A. 7 cm3 of air in a balloon
B. 1 mL of water in a test tube
O C. 4 centimeter piece of string
D. 8 kilograms of bananas
4- Calculate the mol fraction of ethanol and water in
a sample of rectified spirit which contains 95% of
ethanol by mass.
Answer:
math si hard
Explanation:
The mole fration of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass is 0.8 and 0.11.
How do we calculate mole fraction?Mole fraction of any substance will be calculated by dividing the moles of that substance from the total moles of the solution.
Moles (n) will be calculated as:
n = W/M, where
W = given mass
M = molar mass
Given that 95% of ethanol by mass is present in the sample, so 95g of ethanol is present in 100g of solution.
Mass of solvent (water) = 100 - 95 = 5g
Moles of water = 5g / 18g/mol = 0.27mol
Moles of ethanol = 95g / 46g/mol = 2.06mol
Mole fraction of water = 0.27 / (0.27+2.06) = 0.11
Mole fraction of ethanol = 2.06 / (0.27+2.06) = 0.8
Hence required mole fraction of water & ethanol is 00.11 and 0.8 respectively.
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4. Do you think the distribution of elements 10 billion years from now
will be different than it is today? Why or why not?
Answer:
yes
Explanation:
everything will change
Sort each of the following elements by effective nuclear charge, Zeff, from smallest to largest:
Sn, Rb, Sb, In, and Sr
Answer:
Rb = +1 , Sr = +2, In= +3, Sn = +4, Sb= +5
Explanation:
Formula:
Zeff = Z - S
Z = atomic number
S = number of core shell or inner shell electrons
For Sn:
Electronic configuration:
Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²
Zeff = Z - S
Zeff = 50 - 46
Zeff = +4
For Rb:
Electronic configuration:
Rb₃₇ = [Kr] 5s¹
Zeff = Z - S
Zeff = 37 - 36
Zeff = +1
For Sb:
Electronic configuration:
Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³
Zeff = Z - S
Zeff = 51 - 46
Zeff = +5
For In:
Electronic configuration:
In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹
Zeff = Z - S
Zeff = 49 - 46
Zeff = +3
For Sr:
Electronic configuration:
Sr₃₈= [Kr] 5s²
Zeff = Z - S
Zeff = 38 - 36
Zeff = +2
Question 14
4 pts
Using the formula Na + Cl2 --> Naci, if 2.98 moles of sodium are combined with
excess" chlorine, how many moles of sodium chloride will be made? Note that in
this question, excess means extra or as much as you need.
Answer:
2.98 moles of NaCl are produced.
Explanation:
Given data:
Moles of sodium = 2.98 mol
Amount of chlorine = excess
Moles of NaCl produced = ?
Solution:
Chemical equation:
2Na + Cl₂ → 2NaCl
Now we will compare the moles of sodium chloride with only sodium because sodium is limiting reactant and chlorine is in excess thus sodium limit the yield of product.
Na : NaCl
2 : 2
2.98 : 2.98
2.98 moles of NaCl are produced.
Question 4
Volume is what type of physical property?
What ionic substance is being used to illustrate the concept of lattice energy in Model 1?
Answer:
magnesium oxide
Explanation:
hope this helps
How is volume calculated given mass and density? (4 points)
A. Sum of mass and density
B. Mass divided by density
C. Mass multiplied by density
D. Difference of mass and density
12 PTS!! WILL GIVE BRAINLIEST!! DUE TODAY!!
The experiment is increasing and decreasing the amount of predators, pollution, and food to see how the cricket frogs will react.
Test variable (independent variable):
Outcome variable (dependent variable):
Control group:
Answer:
Test variable (independent variable): Predators, Pollution, Food.Outcome variable (dependent variable): Cricket Frogs.Control group: Beginning Frog Count
Explanation:
If pollution isincreased, then the cricket frog population will (stay the same) over the span offive years.If the amount of food available isincreased, then the cricket frog population will (increase) overthe span of five years.Procedure:The procedures are listed in your virtual lab. You do not need to repeat them here. Please be sureto identify the test variable (independent variable), outcome variable (dependent variable), andcontrol group for this experiment.
Ag3PO4 what is the compound with this formula
Answer:
silver (III) phosphate
Explanation:
How many elements are in a isotope?
Answer:i believe 9-?
Explanation:
boron-10
boron-11
carbon-12
carbon-13
oxygen-16
oxygen-17
oxygen-18
chlorine-35
chlorine-37
Kinds of matter with label and example
Answer:
Liquid: water
Solid: Rocks
Gas: Air
Explanation:
are there more metals or nonmetals on the periodic table?
Answer:yes
Explanation:
I think so
There are more metals than nonmetals on the periodic table. The majority of elements on the periodic table are metals, which are found on the left and middle sections of the periodic table.
Metals typically have properties such as high electrical and thermal conductivity, luster, malleability, and ductility.
Nonmetals, on the other hand, are found on the right-hand side of the periodic table, along with the noble gases. Nonmetals generally have properties opposite to those of metals, such as poor electrical and thermal conductivity and a lack of metallic luster. There are fewer nonmetals compared to metals on the periodic table.
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