Answer:
[tex]\rm MgBr_2 \cdot 6\, H_2O[/tex].
Explanation:
Let [tex]x[/tex] denote the number of [tex]\rm H_{2}O[/tex] formula units for every [tex]\rm MgCl_{2}[/tex] formula unit. The formula of the hydrate would be [tex]{\rm MgBr_2} \cdot x\, {\rm H_2O}[/tex].
Look up the relative atomic mass of each element on a modern periodic table:
[tex]\rm Mg[/tex]: [tex]24.305[/tex].[tex]\rm Br[/tex]: [tex]79.904[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].Formula mass:
[tex]\begin{aligned}M({\rm MgBr_{2}}) &= 24.305 + 2 \times 79.904 \\ &= 184.113\; \rm g\cdot mol^{-1}\end{aligned}[/tex].
[tex]\begin{aligned}M({\rm H_{2}O}) &= 2 \times 1.008 + 15.999 \\ &= 18.015\; \rm g\cdot mol^{-1}\end{aligned}[/tex].
The anhydrous compound would be [tex]\rm MgBr_{2}[/tex]. There was [tex]27.21\; \rm g[/tex] of this compound. Calculate the number of moles of formula units in that much of this compound:
[tex]\begin{aligned}n({\rm MgBr_{2}}) &= \frac{m({\rm MgBr_{2}})}{M({\rm MgBr_{2}})} \\ &= \frac{27.21\; \rm g}{184.113\; \rm g \cdot mol^{-1}} \\ &\approx 0.1478\; \rm mol\end{aligned}[/tex].
During heating, [tex]43.19\; \rm g - 27.21\; \rm g = 15.98\; \rm g[/tex] of water [tex]\rm H_{2}O[/tex] escaped from the hydrate. Calculate the number of formula units in that [tex]15.98\; \rm g[/tex] of [tex]\rm H_{2}O\![/tex]:
[tex]\begin{aligned}n({\rm H_{2}O}) &= \frac{m({\rm H_{2}O})}{M({\rm H_{2}O})} \\ &=\frac{43.19\; \rm g - 27.21\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\&= \frac{15.98\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\ &\approx 0.887\; \rm mol\end{aligned}[/tex].
In other words, approximately [tex]0.1478\; \rm mol[/tex] of [tex]\rm MgBr_{2}[/tex] formula units (and hence the same number of moles of [tex]{\rm MgBr_2} \cdot x\, {\rm H_2O}[/tex] formula units) contain approximately [tex]0.887\; \rm mol[/tex] of [tex]\rm H_{2}O[/tex] formula units. Calculate the ratio between the two:
[tex]\begin{aligned}x &= \frac{n({\rm H_{2}O})}{n({\rm MgBr_2})} \\ &\approx \frac{0.1478\; \rm mol}{0.887\; \rm mol} \approx 6\end{aligned}[/tex].
Hence, the formula of the hydrate would be [tex]\rm MgBr_2 \cdot 6\, H_2O[/tex].
How many grams of H2Oare produced when 5.38g of O2
is reacted?
Answer:
I think this answer should be 2.33 g H2O
How is thermal energy distributed in most heating systems?
A.
convection
B.
conduction
C.
radiation
Answer:
By Convection
Explanation:
2.0 kilograms of lithium hydroxide can “scrub”, or remove, how many liters of carbon dioxide at standard conditions?
Answer:
0.94 L
Explanation:
The equation of the reaction is;
2LiOH + CO2 -----> Li2CO3 + H2O
Number of moles LiOH = 2.0/23.95 g/mol = 0.084 moles
2 moles of LiOH scrubs 22.4 L of CO2
0.084 moles of LiOH scrubs 0.084 moles × 22.4/2
= 0.94 L
A nut cracker is second lever. Give reason.
Answer:
the load is between the fulcrum and the handle
Explanation:
A piece of calcium metal was reacted with excess dilute acid to produce hydrogen gas. The gas was collected at SLC. If 4.58g of calcium reacted, what volume of hydrogen gas will be collected?
10g-4.58=4.42g volume of hydrogen gas
A gas that was heated to 150 Celsius has a new volume of 1587.4 L. What was its volume when its temperature was 100 Celsius?
Answer:
[tex]\boxed {\boxed {\sf 1058.3 \ L}}[/tex]
Explanation:
We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:
[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]
The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.
[tex]\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{T_2}[/tex]
The temperature was 100 degrees Celsius, but the volume is unknown.
[tex]\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C}[/tex]
We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.
[tex]100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C} * 100 \textdegree C[/tex]
[tex]100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = V_2[/tex]
The units of degrees Celsius cancel.
[tex]100 *\frac {1587.4 \ L }{150 } = V_2[/tex]
[tex]100 *10.58266667 \ L = V_2[/tex]
[tex]1058.266667 \ L = V_2[/tex]
The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.
[tex]1058.3 \ L = V_2[/tex]
The volume of the gas at 100 degrees Celsius is approximately 1058.3 liters.
The Clean Air Act legislation of 1970, 1977, and 1990 was designed to improve air quality in the United States by monitoring and reducing the emissions of air pollutants judged to pose threats to human health, such as carbon monoxide, nitrogen dioxide, sulfur dioxide, ozone, particulate matter, and lead. The main source of lead emissions in 1970 was the exhaust of vehicles burning gasoline to which tetra-ethyl lead had been added to improve combustion. By 1985, leaded gasoline was phased out of use, although airplanes and racecars were exempted. The 1990 amendments addressed the growing problem of urban smog by
Answer:
The 1990 amendments addressed the growing problem of urban smog by acid rain.
Explanation:
As shown in the above question, the Clean Air Act was created with the goal of making the US air healthier, while also aiming to reduce the impacts of pollution on the American population. This law had as its main objective to reduce the emission of gases capable of polluting the atmosphere, especially those resulting from the burning of gasoline in cars. In 1990, a new amendment was added to this law. This amendment aimed to reduce emissions of sulfur dioxide and nitrogen oxides, as they were the main cause of acid pollution, which was responsible for acid rain that is extremely dangerous and negative for society.
If the [OH1-] of a solution at 25oC is 5.8x10^(-5) mol/L, calculate the [H1+] in mol/L, the pH, and the pOH.
Answer:
1. [H⁺] = 1.58×10¯¹⁰ mol/L
2. pH = 9.8
3. pOH = 4.2
Explanation:
We'll begin by calculating the pOH of the solution. This can be obtained as follow:
Concentration of Hydroxide ion [OH¯] = 5.8×10¯⁵ mol/L
pOH =?
pOH = –Log [OH¯]
pOH = –Log 5.8×10¯⁵
pOH = 4.2
Next, we shall determine the pH of the solution. This can be obtained as follow:
pOH = 4.2
pH =?
pH + pOH = 14
pH + 4.2 = 14
Collect like terms
pH = 14 – 4.2
pH = 9.8
Finally, we shall determine the concentration of Hydrogen ion [H⁺] in the solution. This can be obtained as follow:
pH = 9.8
Concentration of Hydrogen ion [H⁺] =?
pH = –Log [H⁺]
9.8 = –Log [H⁺]
Multiply through by –1
–9.8 = Log [H⁺]
Take the anti log of –9.8
[H⁺] = antilog (–9.8)
[H⁺] = 1.58×10¯¹⁰ mol/L
SUMMARY:
1. [H⁺] = 1.58×10¯¹⁰ mol/L
2. pH = 9.8
3. pOH = 4.2
Plz Help I have One more time To retake and I can't fail it..
Answer:
D
Explanation:
A synthesis reaction is a reaction in which two or more reactants combine to form a given product.
If we look closely at option D, you will notice that sulphur and oxygen combined together to form sulphur dioxide. That is a typical example of chemical synthesis.
Hence;
S + O2 ----->SO2 is a synthesis reaction.
What is nascent hydrogen?
Answer:
[H]
Explanation:
This is nascent hydrogen it's a reactive form of hydrogen...
It reduces elements and compounds.
How many atoms of hydrogen are there in 36 g of NH4?
Answer:
hope it helps you
Explanation:
please like and mark me brainlist
Calcular la m y M de una concentración Porcentual 45% m/m, la cual tiene una D solución 1.25 g/ml y una masa de sto de 125g.HN03
Explanation:
Calculate the "m" and "M" of a percentage concentration of 45% m / m, which has a solution D 1.25 g / ml and a mass of 125g.
Given data is:
The mass % of the solution is 45%.
The density of the solution is 1.25 g/mL.
The mass of the solution is 125 g.
Molarity of solution is:
[tex]M=\frac{number of moles of solute}{volume of solution}[/tex]
The volume of the solution is:
[tex]Volume=\frac{mass}{density} \\=125 g / 1.25 g/ml\\=100 mL[/tex]
From mass% of the solution, the mass of solute HNO_3 can be calculated.
[tex]mass percent=\frac{mass of solute}{ mass of solution} \\45=(mass of HNO_3/125 g)x100\\\\mass of solute = 56.25g[/tex]
The number of moles of HNO3 is :
[tex]n_H_N_O__3=\frac{m_H_N_O__3}{M_H_N_O__3}[/tex]
[tex]=56.25 g/63.00g/mol\\\\=0.893mol[/tex]
The molarity is:
[tex]M=n/V\\M=0.893 mol/0.1L\\M=8.93 M[/tex]
The molality is:
[tex]m=n/mass of solvent in kg\\ =0.893 mol /(0.125-0.05625)kg\\ = 12.9 m[/tex]
. The method of spreading the seeds manually in the leveled ground for crops to grow is called
Answer:
The process of sowing seeds manually called broadcasting or manual sowing. Broadcasting is a method of sowing by scattering seeds all over the prepared soil surface. Crop seeds like wheat, paddy, methi, coriander etc. are sawn in this method.
hii pls help me!!
even if u know how to do one, it's okk
anything helps
Explanation:
a) HNO2(aq) = HNO3(aq) + H2O(l) +NO(g)
b) SoCl2 (l) + H2O (l) = So2(g) + 2HCl(aq)
c) CH4 (g) + 2O2(g) = Co2 (g) + 2H2O(g)
d) 3CuO(s) + 2NH3 (g) = 3Cu(s) + 3H2O (l) + N2(g)
What has an atomic number of 1
Answer:
Hydrogen
Explanation:
When it’s summer in the northern hemisphere what Season is it in the Southern Hemisphere
Answer:
Winter
Explanation:
I dont know but whatev
Answer:
Winter
Explanation:
The seasonal cycle in the polar and temperate zones of one hemisphere is opposite to that of the other. Therefore, When it is summer in the Northern Hemisphere, it is winter in the Southern, and vice versa.
hope it helps :)
What must scientists look for in their experiment or their data to make sure it is accurate?
Answer:
Scientists must look in their experiment/data to check for accuracy because they don't want to make any mistakes, and most importantly have to make sure they have the right information to proceed.
Explanation:
^
PLEASE HELPP!!!
PLEASE HELP!!!
Write the sentences in your copybook and draw a line through one of the words in
bold to complete each of these sentences about alkali metals correctly.
Alkali metals generally become more / less dense going down the group.
The melting and boiling points of alkali metals increase / decrease down the group.
The softness of alkali metals increases / decreases going down the group.
The speed with which alkali metals react with oxygen increases / decreases going
down the group.
Answer:
vahshahhsjajaiaoakkakakaaosos
the number of atoms in 2 moles of aluminium is if the 1mole is 6.022*10 the power of 23 or 10^23
Answer:
1.204 × 10²⁴ atoms
Explanation:
According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.
If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum
2 × 6.02 × 10²3
= 12.04 × 10²³
= 1.204 × 10²⁴ atoms.
Number of sodium atom(s) in salt (NaCl)
Answer:
NaCl consists of one atom each of sodium and chlorine. Hence, each molecule of NaCl has 2 atoms total.
Explanation:
Answer: There is only one atom of sodium in NACL
Explanation: But in total there two atoms because 1 sodium and 1 chlorine added all together
HOPE THIS HELPS!!!!
The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 75 atm at
the same temperature? (5 points)
1) 1.0 L
0 2 2) 1.5 L
3) 2.0 L
O 4) 3.0L
The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at
the same temperature? (5 points)
1) 1.0 L
2) 1.5 L
3) 2.0 L
4) 3.0L
Answer:
2.0 L
Option C) 2.0L is the correct answer.
Hence, at the same temperature, as the pressure is increased to the given value, the volume of the gas decreases to 2.0L.
Boyle's law
Boyle's law states that "the volume V of a given quantity of gas is inversely proportional to its pressure P as long as temperature remains constant.
Boyle's law is expressed as;
P₁V₁ = P₂V₂
Given the data in the question;
Initial Volume V₁ = 2.5LInitial pressure P₁ = 6.0atmFinal pressure P₂ = 7.5atmFinal volume V₂ = ?To determine the final volume of the gas, we substitute our given values into the expression above.
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = ( 6.0atm × 2.5L ) / 7.5atm
V₂ = 15Latm / 7.5atm
V₂ = 2.0L
Option C) 2.0L is the correct answer.
Hence, at the same temperature, as the pressure is increased to the given value, the volume of the gas decreases to 2.0L.
Learn more about Boyle's law here: brainly.com/question/1437490
I need help with question 5
Answer:
B a spring being stretched
what is the charge on ion X Li2X
In the compound Li2X, there are two lithium ions and one X^2- ion.
Ionic compounds are composed of an ion pair of opposite charge. Usually, the positive ion is a metal cation while the negative ion is a non metal anion. The two ions are held together by strong electrostatic interaction.
In the compound Li2X, there are two lithium ions and one X^2- ions. X^2 -is the non metal anion present.
Learn more: https://brainly.com/question/2673886
If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20 degrees C, what is the final temperature of the iron in kelvins?
295.8k
that is the procedure above
The carbon atom forms a part of all the major molecules found in living things. Which of the following doesnot contain carbon?
Hydrochloric acid or Hcl
There is a series of nitrogen oxides with the general formula N?O?. What is the empirical formula of one that contains 30.45% nitrogen?
Question 13 options:
N2O
NO
NO2
N2O3
N2O5
Answer: The empirical formula of one that contains 30.45% nitrogen is [tex]NO_{2}[/tex].
Explanation:
Given: Mass of nitrogen = 30.45 g
Let us assume that the mass of given oxide is 100 grams.
As the atomic mass of nitrogen is 14.0067 g. So, moles of nitrogen will be calculated as follows.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{30.45 g}{14.0067 g/mol}\\= 2.17 mol[/tex]
Also, mass of oxygen = (100 - 30.45) g = 69.55 g
Atomic mass of oxygen is 15.9994 g/mol. So, moles of oxygen will be as follows.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{69.55 g}{15.9994 g/mol}\\= 4.34 mol[/tex]
The ratio of both the atoms is as follows.
[tex]\frac{4.34}{2.17} = 2[/tex]
This means that gas has 2 moles of oxygen to 1 mole of nitrogen. Hence, the formula of oxide is [tex]NO_{2}[/tex].
Thus, we can conclude that the empirical formula of one that contains 30.45% nitrogen is [tex]NO_{2}[/tex].
Combustion reactions typically involve all of the following except
a. The consumption of oxygen gas
b. The production of a Hydrocarbon
c. The production of water
d. The release of carbon dioxide
Answer:
b. The production of a HydrocarbonExplanation:
In Combustion reaction usually occurs when a hydrocarbon reacts with oxygen to produce carbon dioxide and water.
Which molecule will have the strongest LDF
Answer: Try to get answer in Explanation
Explanation: Kr has a greater number of polarizable electrons (36 electrons), so Kr would have a greater degree of LDFs and therefore a greater melting point than N 2 (which has 7 × 2 = 14 polarizable electrons).
How much of a sample remains after five half-lives have occurred?
1/5 of the original sample
1/25 of the original sample
1/32 of the original gample
1/64 of the original sample
Answer:
1/32
Explanation:
Consider an experimental run at 273 K where the initial number of moles (n1) is actually 1.00 mol, and the final number of moles (n2) is 2.00 mol. Use the simulation to find the volume (V1) of 1.00 mol of helium at 273 K, and calculate the final volume (V2).
Express the volume to three significant figures, and include the appropriate units.
Answer: The volume [tex](V_{1})[/tex] of 1.00 mol of helium at 273 K is 22.4 L and the final volume [tex](V_{2})[/tex] is 44.8 L.
Explanation:
Given: [tex]T_{1}[/tex] = 273 K, [tex]n_{1}[/tex] = 1.00 mol
[tex]T_{2}[/tex] = 273 K, [tex]n_{2}[/tex] = 2.00 mol
At the standard pressure, 1 atm the value of [tex]V_{1}[/tex] and [tex]V_{2}[/tex] is calculated as follows.
[tex]V_{1} = \frac{n_{1}RT_{1}}{P}\\= \frac{1.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 22.4 L[/tex]
Similarly,
[tex]V_{1} = \frac{n_{2}RT_{2}}{P}\\= \frac{2.00 mol \times 0.0821 Latm/mol K \times 273 K}{1 atm}\\= 44.8 L[/tex]
Thus, we can conclude that the volume [tex](V_{1})[/tex] of 1.00 mol of helium at 273 K is 22.4 L and the final volume [tex](V_{2})[/tex] is 44.8 L.
