Answer:
By using the criss cross method,the -1 charge of nitrate ion is shifted to ammonium ion and +1 charge of ammonium ion is shifted to nitrate ion. In this way, the final formula NH4NO3 is formed for the ammonium nitrate.
The volume of a gas was 48 mL when the temperature was 159.6 ºC. If the temperature was initially 4.9 ºC, and there was no change in the pressure, what was the initial volume of the gas?
According to Charles law
[tex]\boxed{\sf \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}[/tex]
[tex]\\ \sf\longmapsto \dfrac{48}{159.6}=\dfrac{V_2}{4.9}[/tex]
[tex]\\ \sf\longmapsto V_2=\dfrac{48\times 4.9}{159.6}[/tex]
[tex]\\ \sf\longmapsto V_2=\dfrac{235.2}{159.6}[/tex]
[tex]\\ \sf\longmapsto V_2=1.4mL[/tex]
The unit of force newton is a derived unit.Why?
Answer:
Why is the unit of force called a derived unit? That's because the Newton is defined in terms of other units which are more fundamental. One Newton is the force required to accelerate a mass of one kilogramme at the rate of one metre per second squared.
Identify any three acids.
Identify any three bases.
Balance in basic solution: O2(g) + Cr³+ (aq) → H₂O2 (1) + Cr₂O7²- (aq)
cr2o72+H2o2-(r3+o2)
Explanation:
r62o72+H2o2-(r3+o2)
solution is an example of______mixtures
Answer:
homogeneous mixture must be the answer
What is the freezing point in °C) of a 0.743 m
aqueous solution of KCI?
Enter your rounded answer with
3 decimal places.
K; for water = 1.86 °C/m
Answer:
2.764 °C
Explanation:
ΔT = 2 x 1.86 C/m x 0.743m = 2.764C That means the freezing point of the solution is 2.764C less than the pure solvent (water)
PLZ HELP ME Which of the following is the process by which chloroplasts use carbon dioxide, water, and sunlight to produce sugars and oxygen?
A. Photosynthesis
B. Cellular respiration
c. Reproduction
D. Homeostasis
Answer:
a
Explanation:
what is the most appropriate unit to measure the following; the length of a football field. b) the diameter of a tennis ball
Explanation:
a) m (metre)
b) mm (millimeter)
How much heat is added if .0948g of water is increased in temperature by .728 degrees C?
.289 J
.803 J
1.46 J
.055 J
Answer:
0.289J of heat are added
Explanation:
We can relate the change in heat of a substance with its increasing in temperature using the equation:
q = m*ΔT*S
Where Q is change in heat
m is mass of substance (In this case, 0.0948g of water)
ΔT = 0.728°C
S is specific heat (For water, 4.184J/g°C)
Replacing:
q = 0.0948g*0.728°C*4.184J/g°C
q = 0.289J of heat are added
what is the balanced equation of 2HCl + Mg - MgCla + Ha
2HCl + Mg => MgCl2 + H2
Which of the following is an example of a chemical change?
A. Sugar breaking down into water and carbon when it is heated
B. A sugar cube being crushed by a spoon
C. Sugar mixing with red food coloring and forming a red solid
D. A sugar cube dissolving in water as the water is stirred
Answer:
I think its sugar breaking down into watee and carbon when it is heated
I HOPE ITS RIGHT IF NOT THEN SORRYHAVE A GREAT DAY :)
calcium hydrogen sulphite cress cross
Answer:
Ca Hso4 ×
Explanation:
Because i did whT you said
Formation of HCl from H₂and Cl₂ is example of
photochemical
electrochemical
displacement
Answer:
Combination reaction.
Explanation:
The reaction depicted in the question is;
H2 (g) + Cl2 (g) ⟶ 2HCl (l)
Now, this reaction shows that two reactants namely hydrogen and chlorine react to form a single product named Hydrogen chloride. This process is called combination reaction.
Define physical and chemical properties, provide three examples of each, discuss their reversibility, and explain the fundamental differences between them.
Answer:
Physical properties are defined as the properties which can be observed without changing their chemical composition.
For example, color, volume, and molecular weight.
Chemical properties can be defined as, a properties that can be seen only after changing the substance's chemical identity.
For example, reactivity, toxicity, and flammability.
The fundamental differences between physical and chemical properties are as follows:
Chemical properties are related to chemical bonds of the substance, while physical properties are not.
In chemical properties, the chemical identity of substance changes while physical properties does not have any change.
Chemical properties predict the reaction of a substance, while physical properties only describe the appearance of the substance.
Explanation:
HF or HCl which can form the hydrogen bond? Explain your answer.
Answer:
can only form one hydrogen bond each
Explanation:
can only form one hydrogen bond each
Oxit bazơ có những tính chất hóa học nào?
Answer:
tác dụng với acid,
Explanation:
WILL MARK BRAINLY IF GOOD ANSWER :)
Chlorine and oxygen are both gases at room temperature. What are some characteristic properties of chlorine that distinguish it from oxygen?
Answer:
Well one characteristic is electical energy transforms into thermal energy and gases and the state of matter(one of the distinct form i which matter exist)
Which reaction occurs at the cathode in an electrochemical cell?
A. Cu(s) → Cu2+(aq) + 2e
B. Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)
C. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
D. Cu2+(aq) + 2e → Cu(s)
Answer:
I could be wrong but I say it's letter b
Explanation:
I hope you have an great night
Which of the following has maximum number of atoms?
(a) 18 g of H2O
(b) 18 g of O2
(c) 18 g of CO2
(d) 18 g of CO4
Answer:
(a) 18 g of H2O
Explanation:
The molecule with great number of moles will have great number of atoms.
By observation, since all compounds have same mass, water has the least molcular mass
[tex]H_{2} O = \: 18 \: g[/tex]
Answer:
18g of H2O
Explanation:
the molecule with great number of moles will have great number of atoms .
(a) option is correct
2 Figure 3.1 shows two different forms of carbon, A and B.
A
B
Figure 3.1
a Name the two forms of carbon shown.
Answer:
show the pictures pls specify
Which of the following is the formula for vanadium (II) oxide? Question 11 options: A) VO2 B) VO C) V2O3 D) V2O5
Answer: A. ( VO2)
Explanation:
Answer:
B) VO
Vanadium(II) oxide is the inorganic compound with the idealized formula VO.
1125 J of energy is used to heat 250 g of iron to 55 °C. The specific heat capacity of iron is 0.45 J/(g·°C).
What was the temperature of the iron before it was heated?
55 °C
55 °C
35 °C
35 °C
45 °C
45 °C
20 °C
Answer:
45 °C.
Explanation:
From the question given above, the following data were obtained:
Heat (Q) = 1125 J
Mass (M) = 250 g
Final temperature (T₂) = 55 °C
Specific heat capacity (C) = 0.45 J/gºC
Initial temperature (T₁) =?
The initial temperature of the iron can be obtained as illustrated below:
Q = MC(T₂ – T₁)
1125 = 250 × 0.45 (55 – T₁)
1125 = 112.5 (55 – T₁)
Divide both side by 112.5
1125/112.5 = 55 – T₁
10 = 55 – T₁
Collect like terms
10 – 55 = –T₁
–45 = –T₁
Multiply through by –1
45 = T₁
T₁ = 45 °C
Therefore, the initial temperature of the iron is 45 °C
Which of the following materials contains a base?
a. Stomach digestive juices
b. Orange juice
c. Ammonia window cleaner
d. Vinegar
Answer:
c. ammonia window cleaner
Explanation:
Window cleaners contain ammonia (a base) for the purpose of removing the dust and grease on the windows.
digestive juices, orange(a citrus) juice, and vinegar are all highly acidic.
... give me brainliest please :)) ...
What is the pressure of 4 moles of helium in a 50 L tank at 308 K?
Use PV = nRT.
A. 24.64 atm
B. 0.13 atm
O C. 0.51 atm
D. 2.02 atm
Answer:
D
Explanation:
P = Pressure
V = Volume
n = Moles
R = .08206 (because we're using liters)
T = Temperature (as K)
P(50) = (4)(.08206)(308)
P(50) = 101.09792
Divide both sides by 50
P(50)/50 = 101.09792/50
P = 2.0219584
Answer:
[tex]\boxed {\boxed {\sf D. \ 2.02 \ atm}}[/tex]
Explanation:
We are asked to find the pressure of helium gas using the ideal gas law.
[tex]PV= nRT[/tex]
In this formula P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature.
The pressure is unknown and we are solving for it. The volume is 50 liters, there are 4 moles of helium gas, the universal gas constant is 0.08206 L *atm/ mol * K, and the temperature is 308 Kelvin.
[tex]\bullet \ V= 50 \ L \\\bullet \ n= 4 \ mol \\\bullet \ R = \frac {0.08206 \ L * atm}{mol *K}\\\bullet \ T= 308 \ K[/tex]
Substitute these values into the formula.
[tex]P * 50 \ L= 4 \ mol * \frac {0.08206 \ L * atm}{mol *K} * 308 \ K[/tex]
Multiply on the right side. The units of moles (mol) and Kelvin (K) cancel each other out.
[tex]P* 50 \ L = 4 * 0.08206 \ L *atm * 308[/tex]
[tex]P*50 \ L =0.32824 \ L * atm *308[/tex]
[tex]P* 50 \ L = 101.09792 \ L * atm[/tex]
We are solving for the pressure, so we must isolate the variable P. It is being multiplied by 50 liters. The inverse operation of multiplication is division, so divide both sides by 50 L.
[tex]\frac {P* 50 \ L}{50 \ L} = \frac {101.09792 \ L * atm}{50 \ L}[/tex]
[tex]P= \frac {101.09792 \ L * atm}{50 \ L}[/tex]
The units of liters (L) cancel.
[tex]P= \frac{101.09792}{50 } \ atm[/tex]
[tex]P= 2.0219584 \ atm[/tex]
If we round to the nearest hundredth place the 1 in the thousandth place tells us to leave the 2 in the hundredth place.
[tex]P \approx 2.02 \ atm[/tex]
The pressure is approximately 2.02 atmospheres and choice D is correct.
calculate the maximum mass of ethanol that could be obtained from 30.0g of glucose
Answer:
15.3g
Explanation:
C6H12O6 → 2C2H5OH + 2CO2
1 mol C6H12O6 = 180g
1 mol CH3-CH2-OH = 46g
From 180g glucose ethanol can get = 92g
30g glucose ethanol gives us = (92*30)/180 = 15.33g
aluminum hydroxide molecular weight step by ste
Doing my summer school work right now. Please help me pass I don’t want to take chemistry for the third time.
Answer:
13.B 0.15 M KBr
14.C 1 M C²h⁵OH
15.D 0.1 M C⁶H¹²o⁶
16.A 2 K² SO⁴
17.A 2 M K² SO⁴
18.D 0.028 M KBr
19.A 1 M Nal
20.B 0.1 M AICI³
Infrared waves are an example of heat transfer by
A. convection
B. radiation
C. conduction
D. motion of the air
Answer:
convection is my correct answer
Question 2: thanks!!!
A fixed mass of gas has a volume of 92 cm cube and 3 degrees Celsius. What will be its volume at 18 degrees celsius if the pressure remains constant?
Answer:
94.8454
Explanation:
Let volume be V
Let Temperature be T
V1= 92
T1= 3C but to kelvin 273+3= 300K
V2= ?
T2= 18 C but to kelvin 18+273= 291
[tex] \frac{v1}{t1} = \frac{v2}{t2} [/tex]
[tex] \frac{92}{300} = \frac{v2}{291} [/tex]
[tex]v2 \times 300 = 92 \times 291[/tex]
[tex]v2 = \frac{92 \times 291}{300} [/tex]
[tex]v2 = 94.8454[/tex]
