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IgorLugansk [536]

3 years ago

10

BRAINLIEST PLEASE HELP!!!

1 answer:

Natalka [10]3 years ago

5 0

Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.

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Abby fills a graduated cylinder with 3 mL of water. She

dmitriy555 [2]

Answer:

Volume of the rock, V = 2 mL

Explanation:

It is given that,

Water filled in a graduated cylinder is 3 mL

She then drops a small rock into the cylinder and the water level rises to the 5 mL mark

We need to find the volume of the rock.

The volume of the rock is equal to the difference in volume mark. So,

V = 5 mL - 3 mL

V = 2 mL.

3 0

4 years ago

A cylinder of argon gas contains 50.0 L or Ar at 18.4 atm and 'C. How many moles of argon are in the cylinder?

makkiz [27]

<em>A cylinder of argon gas contains 50.0 L of Ar at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?</em>

<em />

Number moles of Argon : 28.03

<h3>Further explanation</h3>

Given

The volume of gas=50 L

P = 18.4 atm

T = 127+273=400 K

Required

moles of Argon

Solution

Use ideal gas Law :

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{18.4\times 50}{0.08205\times 400}\\\\n=28.03$

5 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

Ba(OH)2:_______.

vova2212 [387]

Answer: D

Explanation: Expand this (OH)2 you will get 2O, 2H

Hence 1Ba, 2O, 2H

5 0

3 years ago

How many grams of oxygen are required to burn 0.10mole of c3h8?

Anastaziya [24]

1mol—44g/mol
0.10mol—x
x=0.10*44
x=4.4 g

8 0

3 years ago