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2 years ago

1. Select all the statements about the nucleus of the atom that are correct:

Chemistry

1 answer:

erik [133]2 years ago

8 0

Answer:

Explanation:

1. Select all the statements about the nucleus of the atom that are correct:

Group of answer choices

B. It contains Protons

D. It has a Positive Charge

E. It contains Neutrons

2. An atom of an element with atomic number 50 and mass number 120 contains:

Group of answer choices

B. 50 protons, 50 electrons, and 70 neutrons

3. Which of these statements is false?

Group of answer choices

D. Electrons have the same mass as a proton but have the opposite charge.

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For which purpose do biochemist insert human genes into bacteria

Mumz [18]

Answer: to find a cure for the common cold to produce insulin for diabetics

Explanation: It has helped many people

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2 years ago

CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST

jekas [21]

Explanation:

2. $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

First, we need to find the number of moles of $CO_2$ at 300K and 1.5 atm using the ideal gas law:

$n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}$

$=2.0\:\text{mol}\:CO_2$

Now use the molar ratios to find the number of moles of ethane to produce this much $CO_2$ .

$2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)$

$=1.0\:\text{mol}\:C_2H_6$

Finally, convert this amount to grams using its molar mass:

$1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)$

$=30.1\:g\:C_2H_6$

3. $3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2$

Convert 75 g Zn into moles:

$75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn$

Then use the molar ratios to find the amount of H2 produced.

$1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2$

Now use the ideal gas law $PV=nRT$ to find the volume of H2 produced at 23°C and 4 atm:

$V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}$

$=8.9\:\text L\:H_2$

8 0

2 years ago

PLEASE HELP ME ASAP THANK YOU!

fomenos

Answer:

The activation energy was reached was 10:45 a.m. The additional energy did not affect the reaction.

Explanation:

7 0

3 years ago

Read 2 more answers

Aluminum nitrate reacts with sodium hydroxide to form a precipitate of aluminum hydroxide. What mass of aluminum hydroxide will

lora16 [44]

Answer:

The mass of aluminum hydroxide that will precipitate when 35.5 mL of 0.145 mol/L aqueous aluminum nitrate is mixed with 44.2 mL of 0.215 mol/L aqueous sodium hydroxide is approximately 0.247 grams

Explanation:

The given parameters are;

The volume of 0.145 mol/L aqueous aluminum nitrate in the reaction = 35.5 mL

The volume of 0.215 mol/L aqueous sodium hydroxide in the reaction = 44.2 mL

The balanced chemical equation for the reaction is given as follows;

Al(NO₃)₃ (aq) + 3NaOH (aq) → Al(OH)₃ (s) + 3NaNO₃ (aq)

Therefore;

1 mole of Al(NO₃)₃, reacts with 3 moles of NaOH to form 1 mole Al(OH)₃, and 3 moles of NaNO₃

The number of moles of aqueous aluminum nitrate in the reaction = 35.5/1000 L × 0.145 mol/L = 0.0051475 moles

The number of moles of aqueous sodium hydroxide in the reaction = 44.2/1000 L × 0.215 mol/L = 0.009503 moles

Therefore, 0.009503 moles NaOH will react with 0.009503/3 moles Al(NO₃)₃ to form 0.009503/3 moles of aluminum hydroxide precipitate

The molar mass of aluminum hydroxide = 78 g/mol

Therefore, 0.009503/3 moles of aluminum hydroxide weighs 78 × 0.009503/3 = 0.247078 grams

The mass of aluminum hydroxide that will precipitate out ≈ 0.247 grams.

7 0

2 years ago

A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container

lara31 [8.8K]

Answer:

n (N₂) = 0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container = 2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas = 5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas = 5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

PV =nRT

n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (N₂) = 6.276 atm .L / 24.52 L atm. mol⁻¹ x 299K

n (N₂) = 0.256 mol

* Find the quantity of O₂

Put value in formula (1)

n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (O₂) = 26.6 atm .L / 24.52 L atm. mol⁻¹

n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

n(Total) = n (N₂) + n (O₂)

n (Total) = 0.256 mol + 1.0848 mol

n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

PV =nRT

P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

P = 1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

P = 2.222 atm

8 0

3 years ago