1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-
137 decays over 60 years, how many grams will remain?

Answer:

Explanation:

Oxidation In chemistry is the gain in oxygen, loss of hydrogen and loss of electrons. Oxidation occur when an element with gain oxygen, loss hydrogen or loss electrons, which means the element is oxidized. Compared to reduction which is loss ofoxygen, gain of hydrogen and gain of electrons.

From the the question , Ag( silver) is undergoing oxidation.

Answer:

The correct answer is 596.5 kJ.

Explanation:

The mass of ethanol or C2H5OH mentioned in the question is 20 gm.  

The molar mass of ethanol is 46 g/mol.  

The moles of the compound can be determined by using the formula,  

n = weight of the compound/molar mass

= 20/46 = 0.435 moles

It is mentioned in the question that standard heat of combustion of ethanol is 1372 kJ/mole, that is, one mole of ethanol is producing 1372 kilojoules of energy at the time of combustion.  

Therefore, the energy liberated by completely burning the 20 grams of ethanol is 0.435*1372 = 596.5 kJ.  

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

Let's consider the thermochemical equation for the combustion of ethanol.

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)   ΔH°c = -1372 kJ/mol

We can calculate the heat released by burning 20.0 g of ethanol considering the following relationships.

[tex]20.0 g EtOH \times \frac{1molEtOH}{46.07gEtOH} \times \frac{(-1367kJ)}{1molEtOH} = -593 kJ[/tex]

Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.

Learn more: https://brainly.com/question/2874342

The standard heat of combustion of ethanol, C₂H₅OH, is -1372 kJ/mol ethanol. How much heat (in kJ) would be liberated by completely burning a 20.0 g sample.

Answer:

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

Explanation:

Let the no. of hours worked as tutor be x

Earning as tutor in 1 hour = $7

Earning as tutor in x hour = $7*x = $7x

Let the no. of hours worked as waiter be y

Earning as waiter in 1 hour = $14

Earning as waiter in y hour = $14*y = $14y

Given that total hours worked in the month = 103 hours

therefore, x + y = 103 ----> (1)

Total amount earned in the month = Earning as tutor in x hour + Earning as waiter in y hour = $(7x + 14y)

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

Explanation:

Ag+(aq) + 2NH3 (aq) ⇄ Ag(NH3)2 (aq)

When it comes to question of this sort, the LeChatelier principle should come to mind. The LeChatelier principle states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

increasing the concentration of Ag+

This would lead to an increase in concentration of reactants, the system would annul or oppose this change by moving towrds the right, favouring the forward reaction.

decreasing the concentration of NH3

This would lead to a decrease in concentration of reactants, the system would annul this change by moving to the left, favourin the backward reaction.

increasing the concentration of Ag(NH3)2

This would lead to an increase in concnetration of products, the system would annul or oppose this change by moving towards the left, favouring the backward reaction.

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

Explanation:

methyl orange turns pink in an acidic solution

yellow in basic and orange in neutral

Answer: The substance would be a strong acid.

Explanation:

Methyl orange turns orange when it is added to a medium acid and turns red when it is added to a strong acid. The pH of the acid would be above 4 because it is a strong acid. Methyl orange turns yellow when it is added to a medium base.

Answer:

vacuum

Explanation:

vacuum is the form by which matter cant exist

Answer:

Plum pudding model

Explanation:

Thomson’s model of the atom called: PLUM PUDDING MODEL

Answer:

.76

Explanation:

Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C  

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

Answer:

there are 5 ways this could happen

Autosomal dominant inheritance:  a child recieves a normal gene from one parent and a defective gene from the other parent.

can occur on any of the 22 non-sex chromosomes and have a 50% inheritence rate, gender is not a factor, and disorder differs with inheritance.

examples:  Huntington's disease, neurofibromatosis, achondroplasia, familial hypercholesterolemia

Autosomal recessive inheritance:  both parents carry the defective gene but they are not affected by the disorder.

there is a 25% chance of defective gene from both parents, a 50% chance of inheriting one gene to become a carrier, gender is not a factor in the pattern of the defective gene.

examples:  Tay-Sachs disease, sickle cell anemia, cystic fibrosis, phenylketonuria (PKU)

X-linked (sex-linked) recessive inheritance:  mother carries the affective gene on one of the two X chromosomes.

males inherite X chromosomes from their mothers and Y from their father; which gives the son a 50% chance of inheriting the disorder.

daughters have a 50% chance, but they are not affected by the disorder.

examples:  Hemophilia A, Duchenne muscular dystrophy

X-linked Dominant:  females are affected more so than males; more common for males if they are in the same generation if the mom is affected (because females have two X-chromosomes)

example:  Hypophatemic rickets (Vitiamin Dresistant rickets, ornithine transcarbamylase deficiency.

Mitochondrial:  can affect both males and femlaes, can only be passed by females due to all mitochondria of all children is from the mother, and can appear in every generation.

examples:  Lebrer's hereditary optic neuropathy and Kearns-Sayre syndrome

Explanation:

Answer:

I believe the answer would be: D

an independent variable that is affected by the other

experimental variable

Correct me if im wrong

Explanation:

Answer:

4>3>1>2.

That is, the C-C bond length in ethane > benzene > ethene > ethyne.

Explanation:

The C-C bond in ethane is single, the C-C bond in ethene is double and the C-C bond in ethyne is triple. As the number of bonds between a C-C increases, the length of the bond decreases with an increase in strength. This explains why the C-C bond length in ethane > ethene > ethyne. For benzene, all the C-C bonds in the aromatic compound has been found to have an identical length of 1.40 Å, compared to ethane (1.54Å), ethene (1.34Å) and ethyne (1.20Å). Hence the trend in bond lengths: ethane > benzene > ethene > ethyne.

The question is incomplete; the complete question is;

Which of the following statements is/are true for a 0.10 M solution of a weak acid HA?

a. [ H+] >> [ A−]

b. [ H+] = [ A−]

c. The pH is 1.00.

d. The pH is less than 1.00.

Answer:

b. [ H+] = [ A−]

Explanation:

Given the acid as HA, we know that being a weak acid, its dissociation in water can never be 100%. If it were a strong acid, then it could have undergone a 100% dissociation in solution. The conjugate base of a weak acid is a always a weak base hence A^- is expected to act as a weak base. At the same concentration, weak acids have a higher pH value than strong acids. Hence if the pH of a strong acid HA is 1, then the pH of a weak acid HA must be greater than 1.

But, we look at the equation for the dissociation of the weak acid HA

HA(aq)⇄H^-(aq) + A^-(aq). This implies that the HA dissociates in a 1:1 ratio therefore; [H+] = [ A−], hence the answer given above.

Answer:

Scientific notation is a system in which quantities are too big or too tiny to compose in decimal form.

Key words:

1: Scientific

2: Quantities

3: Decimal

Please mark brainliest

Hope this helps.

Answer:

  2.3

Explanation:

Your calculator can evaluate this expression. It tells you the pH of lemon juice is 2.3.

Answer: The molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Explanation:

We are given:

Mass of [tex]CO_2=4.13g[/tex]

Mass of [tex]H_2O=1.13g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, =[tex]\frac{12}{44}\times 4.13=1.13g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, [tex]\frac{2}{18}\times 1.13=0.125g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles[/tex]

Moles of Hydrogen =[tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles[/tex]

Moles of Oxygen =[tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles

For Carbon = [tex]\frac{0.094}{0.094}=1[/tex]

For Hydrogen = [tex]\frac{0.125}{0.094}=1.33[/tex]

For Oxygen = [tex]\frac{0.109}{0.094}=1.16[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is [tex]C_6H_8O_7[/tex]

Empirical mass = [tex]6\times 12+8\times 1+7\times 16=192g[/tex]

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

Putting values in above equation, we get:

[tex]n=\frac{192g/mol}{192g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_6H_8O_7\times 1=C_6H_{8}O_7[/tex]

Thus molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]

Answer: The right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

[tex]OH^-(aq)+HCO_3^-(aq)\rightleftharpoons H_2O(l)+CO_3^{2-}(aq)[/tex]

Here, [tex]OH^-[/tex] is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms [tex]H_2O[/tex] which is a conjugate acid.

Thus the right conjugate of [tex]OH^-[/tex] is [tex]H_2O[/tex]

Answer:2.2059

Explanation:find their total ram is(14+(1×3))

Find the portion occupied by

H2 which is three

Divide h2 by total ram then multiply by mass

Answer:

The formula is as follows

2Cu + O2 ---> 2CuO

When two copper atoms react with a diatonic oxygen, they form copper oxide (rust)

Answer:

C = 4.60x10⁻⁵ M

Explanation:

The concentration of the compound can be calculated using Beer-Lambert Law:

[tex]A = \epsilon*C*l[/tex]

Where:

A: is the absorbance of the ethanol = 0.58  

ε: is the molar absorptivity of the ethanol = 12600 M⁻¹cm⁻¹

C: is the concentration of the compound =?

l: is the optical path length = 1 cm

Hence, the concentration of the compound is:

[tex]C = \frac{A}{\epsilon*l} = \frac{0.58}{12600 M^{-1}cm^{-1}*1 cm} = 4.60 \cdot 10^{-5} M[/tex]

Therefore, the concentration of the compound is 4.60x10⁻⁵ M.

I hope it helps you!

Answer:

B

Explanation:

Answer:

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

Replacing:

51.0345 atm*0.890 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *294 °K

Solving:

[tex]n=\frac{51.0345 atm*0.890 L}{0.082\frac{atm*L}{mol*K} *294 K}[/tex]

n=1.884 moles

Being 32 g / mole the molar mass of oxygen O₂, the following rule of three applies: if 1 mole contains 32 grams, 1,884 moles, how much mass does it have?

[tex]mass=\frac{1.884 moles*32 grams}{1mole}[/tex]

mass= 60.288 grams

60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.

Answer:

The mass of the O2 gas is 60.16 grams

Explanation:

Step 1: Data given

Volume of O2 gas = 890.0 mL = 0.890 L

Temperature of the O2 gas = 21.0 °C = 294 K

Pressure = 750.0 psi = 51.03 atm

Step 2: Calculate moles of O2 gas

p*V = n*R*T

⇒with p = the pressure of the O2 gas = 51.03 atm

⇒with V = the volume = 0.890 L

⇒with R = the gas constant == 0.08206 L*atm/mol*K

⇒with T = the temperature = 294 K

⇒with n = the number of moles of O2 gas

n = (p*V) / (R*T)

n = (51.03 atm * 0.890 L) / (0.08206L*atm/mol*K * 294K)

n = 1.88 moles of O2

Step 3: Calculate the mass of O2 gas

Mass = moles * molar mass O2

Mass O2 gas = 1.88 moles * 32.0 g/mol

Mass O2 gas = 60.16 grams

The mass of the O2 gas is 60.16 grams

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C[tex]_{V}[/tex] You can work out C[tex]_{V}[/tex]

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC[tex]_{V}[/tex] ∆T

Polyatomic gas: C[tex]_{V}[/tex] = 3R

∆U= nC[tex]_{V}[/tex] ∆T

∆U= 28g x C[tex]_{V}[/tex] x (350K - 58.3K)

∆U = 28C[tex]_{V}[/tex] x 291.7

∆U = 10967.6 x C[tex]_{V}[/tex]

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of  0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

Answer:

[tex]M_{H_3PO_4}=0.157M[/tex]

Explanation:

Hello,

In this case, the balanced chemical reaction is:

[tex]H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)[/tex]

Therefore, we compute the moles of used NaOH by the 0.2000-M solution:

[tex]n_{NaOH}=0.200\frac{mol}{L}*0.0236L=4.72x10^{-3}molNaOH[/tex]

Then, we compute the moles of neutralized phosphoric acid by their 1:3 molar ratio:

[tex]n_{H_3PO_4}=4.72x10^{-3}molNaOH*\frac{1molH_3PO_4}{3molNaOH}=1.57x10^{-3}molH_3PO_4[/tex]

Finally, the concentration:

[tex]M_{H_3PO_4}=\frac{1.57x10^{-3}molH_3PO_4}{0.010L}\\ \\M_{H_3PO_4}=0.157M[/tex]

Best regards.

Answer:

This is true because oxygen belongs to the non-metal part of the periodic table.

Answer:hello

Oxygen is a non-metal it's exactly true.

Explanation:

There's a good rule that if the number of electrons in the last layer is 1,2,3,and sometimes 4 it's a metal but if this number increase it's a non-metal like Oxygen or Nitrogen.

Good luck.

Answer:

Collision theory states that the rate constant for a chemical reaction is composed

of three factors, (1) the absolute number of collisions, Z, between molecules; (2) The

fraction of collisions, f, with an energy greater than the activation energy; and (3) the

fraction of molecules, p, in which the molecules are in the correct orientation to react.

k = Zfp Equation 1

The absolute number of collisions, Z, increases with temperature. However, it has

been shown that at 25o

C, the increase in the number of collisions accompanying a 10o

C

increase in temperature accounts for only about 2% of the increase in the reaction rate.

Similarly, while it is important that molecules be in the proper orientation to react when

they collide, molecular orientation is independent of temperature. Thus it follows that the

major factor controlling reaction rates is the fraction, f, of molecules in the reaction

mixture with an energy greater than the activation energy. This factor, f, depends on the

absolute temperature. It has been shown that f is related to Ea by the following equation

Explanation:

Answer:

The volume of base required is = 25.66 mL

Explanation:

Volume of titrant required is given by:

Final titre value - Initial titre value

=26.78 - 1.12

= 25.66 mL

Note: repeat and average volumes for accuracy of result.