1. 2.50 g of hydrogen gas is heated from 17.0°C to 23.0°C. The specific heat capacity of hydrogen is 14.267 J/g C. How much energy is absorbed?​

Answer: Yes

Explanation:

Duh it can change.  Because matter is what we are made of, and if we change, then matter also changes along with us :)

Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

Best regards!

The lewis structure of given molecule is given below. Carbon has 4 electron. out of which electron make bond with carbon and one with hydrogen and chlorine.

If the compound's molecular formula is known, electron dot structures or Lewis dot formulas can be drawn. It describes the type of bond and the placement of the atoms within the molecule that are joined together. In homage of American chemist Gilbert Newton Lewis, molecules are depicted as having a Lewis electron dot structure or simply a Lewis structure. The lewis structure of given molecule is given below.

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Answer:

of, relating to, or having a strong affinity for water

Explanation:

Answer:

water loving or havin affection for water, not like hydrophobic wich is fear of water or not resistant to water

Explanation:

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

Moles K₂CO₃ = Moles CO₃²⁻:

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

Answer:

9.52 × 10³ J

Explanation:

There is some info missing but I found it on the web.

Calculate the energy required to heat 179g of ethanol from -2.2 °C to 19.6 °C. Assume the specific heat capacity of ethanol under these conditions is 2.44 J/g.°C. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Calculate the temperature change

ΔT = 19.6 °C - (-2.2 °C) = 21.8 °C

Step 3: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 2.44 J/g.°C × 179 g × 21.8 °C

Q = 9.52 × 10³ J

3 × 10²⁴ atoms Cd

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

5 mol Cd (Cadmium)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 1 sig fig.

3.011 × 10²⁴ atoms Cd ≈ 3 × 10²⁴ atoms Cd

Answer:

HCL, NACL and CO are the examples of chemical formulas

Answer:

The answer is 146.

Explanation:

To calculate speed, you have to divide the distance by the time. So, 60/0.41 would be approximately 146.

Answer:

boring

Explanation:

life would be meaningless like everyone being like on person

The question is incomplete, the complete question is;

When 1 mole of P4O10(s) reacts with H2O(l) to form H3PO4(aq) according to the following equation, 453 kJ of energy are evolved.

P4O10(s) + 6H2O(l)When 1 mole of P4O10(s) reacts4H3PO4(aq)

Is this reaction endothermic or exothermic? _________endothermicexothermic

Answer:

The reaction is exothermic

Explanation:

An exothermic reaction refers to any reaction in which heat was EVOLVED. The word was stressed to indicate that it holds the key to answering the question correctly.

We were told in the question that heat was evolved, hence the reaction is exothermic in nature.

The balanced thermochemical equation of the reaction is :

P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) ____ ΔH = -453 kJ

Answer:

1. When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules

2. Pure sodium explodes on contact with water

Explanation:

Answer: The hydroxide concentration of this sample is [tex]3.85\times 10^{-7}[/tex]

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

[tex]K_w=[H^+]\times [OH^-][/tex]

[tex]7.7\times 10^{-14}=[H^+]\times [OH^-][/tex]

As [tex][H^+]=[OH^-][/tex]

[tex]2[OH^-]=7.7\times 10^{-14}[/tex]

[tex][OH^-]=3.85\times 10^{-7}[/tex]

Thus hydroxide concentration of this sample is [tex]3.85\times 10^{-7}[/tex]

It is not D or C

Pick A or B

Answer:

D Flat Top Hills Can Be Plateaus or mesas

Explanation:

Answer:

We need to add 900 mL of water.

Explanation:

To know the amount of water to add first we need to calculate the concentration of KOH when the pH is 12 and when is 11:

[tex] 14 = pH + pOH [/tex]

[tex] pOH = -log[OH] [/tex]

[tex] OH = 10^{-pOH} [/tex]

When pH = 12 (initial):

[tex]pOH_{i} = 14 - 12 = 2[/tex]

[tex]OH_{i} = 10^{-2}=0.01 M[/tex]

When pH = 11 (final):

[tex]pOH_{f} = 14 - 11 = 3[/tex]

[tex]OH_{f} = 10^{-3} = 0.001 M[/tex]

Now, by using the following equation we can find the volume of the solution to achieve a pH = 11:

[tex]C_{i}V_{i} = C_{f}V_{f}[/tex]

[tex]V_{f} = \frac{C_{i}V_{i}}{C_{f}} = \frac{0.01 M*100 mL}{0.001 M} = 1000 mL[/tex]

Finally, to bring 100 mL of the initial solution to 1000 mL we need to add the following amount of water:

[tex]1000 mL = V_{KOH} + V_{H_{2}O}[/tex]

[tex] V_{H_{2}O} = 1000 mL - V_{KOH} [/tex]

[tex] V_{H_{2}O} = 1000 mL - 100 mL = 900 mL [/tex]

Therefore, we need to add 900 mL of water.

I hope it helps you!            

The amount of water that needs to be added will be 900 mL

For finding the amount of water added we have to calculate the concentration of KOH at pH=12 and pH=11

[tex]14=pH+_pOH[/tex]

[tex]_pOH=-Log[OH][/tex]

[tex]OH=10^{-_pOH}[/tex]

[tex]_pOH_f=14-11=3[/tex]

When pH = 12 (initial):

[tex]_pOH_i=14-12=2[/tex]

[tex]_pOH_i=10^{-2}=0.01[/tex]

When pH = 11 (final):

[tex]_pOH_i=10^{-2}=0.01[/tex]

[tex]_pOH_f=14-11=3[/tex]

Now, by using the following equation we can find the volume of the solution to achieve a pH = 11:

[tex]C_iV_i=C_fV_f[/tex]

[tex]V_f=\dfrac{C_iV_I}{C_f} =\dfrac{0.01\times100}{0.001} =1000\ mL[/tex]

Finally, to bring 100 mL of the initial solution to 1000 mL we need to add the following amount of water:

[tex]1000=V_{KOH}+V_{H_2O}[/tex]

[tex]V_{H_2O}=1000mL-V_{KOH}[/tex]

[tex]V_{H_2O}=1000-100=900mL[/tex]

Thus the amount of water that needs to be added will be 900 mL

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7.53 × 10²³ molecules NaOH

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

50.0 g NaOH

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of NaOH - 22.99 + 16.00 + 1.01 = 40 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

7.5275 × 10²³ molecules NaOH ≈ 7.53 × 10²³ molecules NaOH

2 L Ne

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

0.07 mol Ne (g)

Step 2: Identify Conversions

STP - 22.4 L per mole

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 1 sig fig.

1.568 L Ne ≈ 2 L Ne

Answer:

[tex]t=1.712min[/tex]

Explanation:

Hello!

In this case, since the radioactive decay equation is:

[tex]\frac{A}{A_0}=2^{-\frac{t}{t_{1/2} }[/tex]

Whereas A stands for the remaining amount of this sample and A0 the initial one. In such a way, since the sample of rutherfordium is reduced to one-third of its nuclei, the following relationship is used:

[tex]A=\frac{1}{3} A_0[/tex]

And we plug it in to get:

[tex]\frac{\frac{1}{3} A_0}{A_0}=2^{-\frac{t}{t_{1/2}} } \\\\\frac{1}{3}=2^{-\frac{t}{t_{1/2}} }[/tex]

Now, as we know its half-life, we can compute the elapsed time for such loss:

[tex]log(\frac{1}{3})=log(2^{-\frac{t}{t_{1/2}} })\\\\log(\frac{1}{3})=-\frac{t}{t_{1/2}} }*log(2)[/tex]

[tex]t=-\frac{log(\frac{1}{3})t_{1/2}}{log(2)} \\\\t=1.71min[/tex]

Best regards!

Answer:

Xenon

Atomic Number: 54

Explanation:

Periodic Table

Answer:

Xenon

Explanation:

Xenon is a chemical element with symbol Xe and atomic number 54.

Answer:

Please put question in!

Explanation:

Answer:

38.1

Explanation:

8+6+4+3=21

8/21

38.1

Answer:

it is a substance

Explanation:

because sodium is an element which its consistent is only sodium atom

Answer:

C) The law of conservation of matter

Explanation:

Conservation means keeping/ retaining something, in this case it is matter. Hope this helps!

Answer:

it's the 2 I'm pretty sure

Explanation:

please give me a heart

Answer:

Explanation:

sd dsf sd f sdf sdf

Answer:

The temperature of the mixture is 60°C

Explanation:

We can write the energy of water as follows:

E = m×C×T

Where E is energy in Joules, m is mass of water, C is specific heat of water = 4.184J/g°C and T is temperature

Replacing fot both samples:

E = 20g×4.184J/g°C×30°C

E = 2510.4J

E = 40g×4.184J/g°C×75°C

E = 12552J

The total energy of the mixture is 12552J + 2510.4J = 15062.4J

Mass = 60g:

15062.4J = 60g×4.184J/g°C×T

60°C = T

Answer:

V₂ = 70.80 L

Explanation:

Given data:

Initial volume of He = 45 L

Initial temperature = 250 °C

Final temperature = 550 °C

Final volume = ?

Solution:

Initial temperature = 250 °C (250+273.15 K= 523.15 K)

Final temperature = 550 °C (550 + 273.15 k= 823.15 K)

According to Charles Law.,

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 45 L × 823.15  K / 523.15 K

V₂ = 37041.75 L.K / 523.15 K

V₂ = 70.80 L