3. How many grams of water will absorb a total of 4,520 joules of eneray when the temperature
changes from 25°C to 82.7°C

Answers

Answer 1
q = mcΔT

q = 4520 J
ΔT = 82.7 - 25 = 57.7°C
c = 4.182 J/g°C

4520 = m(4.182)(57.7)
m = 18.7 g

18.7 g of water

Related Questions

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)

Answers

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.

Answers

Answer:

[tex]\alpha=17.7[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=13g[/tex]

Volume [tex]V=10mL[/tex]

Angle [tex]\theta=23[/tex]

Sample Tube=10cm

Generally the equation for concentration is mathematically given by

 [tex]C=m/v[/tex]

 [tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]

Therefore the Specific Rotation

 [tex]\alpha=frac{\theta }{m*l}[/tex]

 [tex]\alpha=frac{23 }{1.3*1.0}[/tex]

 [tex]\alpha=17.7[/tex]

chemical formula for Ammonia

Answers

Answer:

NH3

Explanation:

It’s a polyatomic ion. You’ll just have to memorize it!

Answer:

NH3

Explanation:

Ammonia is a compound of nitrogen and hydrogen with the formula NH3

A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.

Answers

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

(c) m X is an ion in which group of the periodic table is the element from which X is formed?​

Answers

Explanation:

Iron has 2 atoms and 3atoms.

So,X=2,3

state the number of protons contained in the atomic number 8

Answers

Answer:

There are 8 protons

Explanation:

[tex]{ \bf{atomic \: number = proton \: number}}[/tex]

How many moles of Al are necessary to form 45.0 g of AlBr, from this
reaction:
2 Al(s) + 3 Br_(1) ► 2 AlBr_(s)?

Answers

Answer:

0.169 mole of Al

Explanation:

We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:

Mass of AlBr₃ = 45 g

Molar mass of AlBr₃ = 27 + (3×80)

= 27 + 240

= 267 g/mol

Mole of AlBr₃ =?

Mole = mass /molar mass

Mole of AlBr₃ = 45 / 267

Mole of AlBr₃ = 0.169 mole

Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:

2Al + 3Br₂ —> 2AlBr₃

From the balanced equation above,

2 moles of Al reacted to produce 2 moles of AlBr₃.

Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.

Thus, 0.169 mole of Al is needed for the reaction.

What force is behind us when we ride a bike?

Answers

Answer:

gravity, ground, friction, rolling resistance, and air resistance.

gravity and force which helps us to not a
fall and keep going

Which particle has a mass of 9.11 x 10^-28g and charge of -1?
A. electron
B. proton
C. neutron​

Answers

QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?

OPTIONS:-

A. electron

B. proton

C. neutron

ANSWER:-

CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER

THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER

MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]

CHARGE ON ELECTRON:- [tex] -1[/tex]

SO URE ANSWER IS ELECTRON

Cho một thực phẩm có độ ẩm tương đối là 81%, hỏi hoạt độ của nước trong thực phẩm đó là

Answers

Given question is:

Given a food with a relative humidity of 81%, what is the water activity in that food?

Explanation:

Water activity in food can be determined by using the below-shown formula:

[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]

Equilibrium is established between the vapor pressure of food and the surrounding air media.

Thus, for the given food the relative humidity is 81%.

hence, its water activity is

[tex]81/100\\=0.81[/tex]

Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction

Answers

Answer:

The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"

Explanation:

Calculating the rate of the equation:

[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]

Rate:

[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]

10.0 g of Mg are reacted with 95.0 g of I2. If 27.5 g of magnesium iodide are obtained, what is the percent yield

Answers

Answer:

% yield of reaction is 26.4

Explanation:

The reaction is:

Mg + I₂ →  MgI₂

Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:

10 g . 1mol / 24.3 g = 0.411 moles of Mg

95 g . 1mol / 253.8g = 0.374 moles of I₂

Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine

For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.

As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂

We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g

To calculate the percent yield:

% yield = (yield produced /theoretical yield) . 100

% yield = (27.5 g/ 104g) . 100 = 26.4

Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g) -->
2Ag2S(s) + 2H2O(g)

Answers

Answer:

Oxygen is being oxidized while Ago is reduced

8moles of Na2Cr2O2 is how much mass​

Answers

[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]

[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]

= 2 × 23 + 2 × 52 + 2 × 16

= 182 grams

1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g

8 moles weigh = 8× 182

=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]

or

[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]

once formed, how are coordinate covalent bonds different from other covalent bonds?

Answers

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

Choose a Metal to Make Cookware
Aluminum
Copper
Iron
Lead
0.90
0.35
0.44
0.12.
Specific heat (J/g°C)
Cost ($ per lb)
1.00
5.00
0.10
1.00
Safety risk
slight
slight
none
significant
Density (g/cm3)
2.70
8.92
7.87
11.30
Considering only specific heat, V lead would
be the most ideal for use in cookware.
COMPLETE
However, compared to the other options, lead is
too

Answers

Answer:

lead

Explanation:

All metals are good conductors of heat and electricity. This property allows the heat to flow through them. The metals which are used to make cookware are Aluminum, Copper and Iron. The correct options are A, B and C.

What are metals?

The metals are defined as the substances which are formed naturally below the surface of the earth. Most of the metals are lustrous or shiny. They are very strong and durable, so they are used to make many things like satellites, automobiles, cooking utensils, etc.

Malleability is the property of metals which allows them to be beaten into flat sheets. Aluminium sheets are used in the manufacture of aircrafts and utensils because of its light weight and strength.

So here aluminium, copper and iron are used to make utensils, since they are good conductors of heat and electricity.

Thus the correct options are A, B and C.

To know more about metals, visit;

https://brainly.com/question/20570629

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In the given range,at what temperature does oxy gen have the highest solubility?​

Answers

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.
25oC
Solubility of oxygen and oxygen compounds

Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.

Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force

Answers

Hydrogen bonding
In the secondary structure of a protein, hydrogen bonds between amino acids determine the configuration of the molecules.
In DNA, hydrogen bonds connect the nitrogenous bases (2 hydrogen bonds between adenine and thymine, 3 hydrogen bonds between guanine and cytosine)

Answer:

hydrogen bonding

Explanation:

just took the test :D

Step 7: Measuring the Volume of Air Near 60°C

Answers

Answer:

57------6.9

330------.87

Explanation:

Answer:

The required volume of air is 3.64 L

Explanation:

Ideal gas equation:-

The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,

   PV=RT

Where R=gas constant

Now,

Let the volume of air is, V

According to the question we have

Temperature, T= 60°C= (60+273) K= 333 K

Atmospheric pressure, P= 760 mm

And gas constant, R= 8.314 J/mole K

Substitute values in ideal gas equation and we get

  760V= 8.314(333)

  V= 2768.562/760

  V= 3.642 L

Learn more about ideal gas here:

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Consider the following reaction:

CO(g)+2H2(g)⇌CH3OH(g)

A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.

Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answers

Answer:

26.6

Explanation:

Step 1: Calculate the molar concentrations

We will use the following expression.

M = mass solute / molar mass solute × liters of solution

[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M

[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M

[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M

Step 2: Make an ICE chart

        CO(g) + 2 H₂(g) ⇄ CH₃OH(g)

I        0.184      0.227           0

C         -x           -2x             +x

E     0.184-x   0.227-2x        x

Since [CH₃OH]e = x, x = 0.0523

Step 3: Calculate all the concentrations at equilibrium

[CO]e = 0.184-x = 0.132 M

[H₂]e = 0.227-2x = 0.122 M

[CH₃OH]e = 0.0523 M

Step 4: Calculate the equilibrium constant (Kc)

Kc = [CH₃OH] / [CO] [H₂]²

Kc = 0.0523 / 0.132 × 0.122² = 26.6

The fuel used in many disposable lighters is liquid butane, C4H10. How many carbon atoms are in 1.00 g of butane

Answers

Answer:

4.15×10²² atoms of C are contained in 1 g of butane.

Explanation:

Formula for butane is C₄H₁₀

1 mol of butane contains 4 moles of carbon and 10 moles of H

Molecular weight is 58 g/mol

In conclussion we can say that 58 g of butane contain 4 moles of C and 10 moles of H

We prepare the rule of three:

1 g . 4 moles of C / 58g = 0.0689 moles of C

Now we can count the atoms by NA. We know that 6.02×10²³  particles are contained in 1 mol of anything.

0.0689 mol . 6.02×10²³ atoms / 1mol = 4.15×10²²

4.15×10²² atoms of C are contained in 1 g of butane.

Predict the reactants of this chemical reaction.

products: KClO4 + H2O
reactants: ?

Answers

Answer:

KOH + HClO4 = K(ClO4) + H2O

Explanation:

KClO4 + H2O : mix and match the letters if you need to guess

How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma

Answers

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

What is the energy change when 78.0 g of Hg melt at −38.8°C

Answers

Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if of methane were consumed

Answers

Answer:

4.4 mL

Explanation:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed?

Step 1: Write the balanced equation

CH₄(g) + 4 Cl₂(g) ⇒ 4 HCl(g) + CCl₄(g)

Step 2: Establish the appropriate volume ratio

For gases under the same conditions, the volume ratio is equal to the molar ratio. The molar ratio of CH₄ to HCl is 1:4.

Step 3: Calculate the volume of HCl produced from 1.1 mL of CH₄

1.1 mL CH₄ × 4 mL HCl/1 mL CH₄ = 4.4 mL HCl

An enzyme increases the reaction rate by:________. a. stabilizing of transition state. b. shifting the reaction equilibrium. c. increasing the probability of product formation. d. All of the answers are correct. e. None of the answers is correct.

Answers

Enzymes provide activation energy for the reactions they catalyze.

Identify the phase of the copper product after each reaction in the copper cycle.

The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )

Answers

Answer:

addition of HNO3 HNOX3 to Cu - Aueous

addition of H2SO4 HX2SOX4 to CuO - Aqueous

addition of Z n Zn to C u S O 4 CuSOX4 - Solid

addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid

heating of C u ( O H ) - Solid

Explanation:

Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.

How can this product be achieved using the starting material shown?

Answers

Answer:

this product can be achieved using the starting material shown is by use of NaOH as catalyst.

Answer:

By using NaOH as catalyst.

Explanation:

This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.

a pure sample of a compound on combustion analysis gave 361 mg of CO2 and 147mg of H2O. if the weight of the sample is 202 mg calculate the weight of carbon in the sample​

Answers

Answer:

[tex]m_C=98.5mgC[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us obtain the mass of carbon in the sample by applying the following stoichiometric factor, considering that the only source of this atom in the sample is contained in the produced 361 mg of CO2 and one mole of this compound contain one mole of carbon atoms whose mass is 12.01 g/mol or mg/mmol as shown below:

[tex]m_C=361mgCO_2*\frac{1mmolCO_2}{44.01mgCO_2}* \frac{1mmolC}{1mmolCO_2}*\frac{12.01mgC}{1mmolC} \\\\m_C=98.5mgC[/tex]

Regards!

Which daughter element is produced from the alpha decay of 213 over 85 At ?
A. 213 over 86 Rn
B.217 over 87 Fr
C. 213 over 84 Po
D. 209 over 83 Bi​

Answers

Answer:

209

83 Bi

Explanation:

213 213 - 4 4

85 At = 85 - 2 Y + 2 He

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