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Allisa [31]
4 years ago
5

A sample of 0.800 mol of nitrogen has a pressure of 0.600 atm and a

Chemistry
1 answer:
schepotkina [342]4 years ago
5 0

Answer:

274K

Explanation:

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Molecules with ____________ bonds between their components will ____________ (fall apart) in water, releasing charged particles
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Molecules with ionic bonds between their components will dissociate (fall apart) in water, releasing charged particles called ions.<span>
<span>When the number of protons and electrons are not equal, ion is formed. There are two types of ions, Cations and anions. Cations are the positively charged ions and when an atom gains electron, Cation is formed. Anions are negatively charged ions and when an atom loses electrons, anion is formed.</span></span>

8 0
3 years ago
2. Calculate how many moles of H2 would be produced if 0.250 mol of Fe reacts completely. (1 mark)
den301095 [7]

Hey there!:

Balanced Equation , for the reaction:

3 H₂SO₄(aq) + 2 Fe(s) → Fe₂(SO₄)₃(aq) + 3 H₂(g)

2 moles Fe ---------------------------------- 3 moles H₂

0.250 moles Fe ---------------------------- ( moles H₂ ) ??

____________________________________________

moles H₂ = 0.250 × 3 / 2

moles H₂ = 0.75 / 2

moles = 0.375 moles of H₂

Hope this helps!

7 0
3 years ago
WHAT IS THE CHEMICAL FORMULA FOR WATER
galina1969 [7]

Answer:

are you a kid bruh xd

Explanation:

H2O

3 0
3 years ago
Read 2 more answers
HELP!!!!! How much does a sample of Argon weigh if it occupies 25.4 L at a pressure of 2.45 atm and a temperature of 482 K?
Artist 52 [7]

Answer:

62.82 g

Explanation:

From the question,

PV = nRT.................. Equation 1

Where P = Pressure, V = Volume, n = number of moles of argon, R = molar gas constant, T = Temperature.

But,

Number of mole (n) = mass (m)/molar mass(m')

n = m/m'................... Equation 2

Substitute equation 2 into equation 1

PV = (m/m')RT.............. Equation 3

From the question, we were asked to find m.

There make m the subject of formula in equation 3

m = PVm'/RT.............. Equation 4

Given: P = 2.45 atm, V = 25.4 L, T = 482 K

Constant: R = 0.082 atm.dm³.K⁻¹.mol⁻¹, m' = 39.9 g/mol

Substitute these values into equation 4

m = (2.45×25.4×39.9)/(0.082×482)

m = 62.82 g.

5 0
3 years ago
Can someone help please?
attashe74 [19]
<h3>Take the weighted average of the individual isotopes.</h3><h3 /><h3>Explanation:</h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>69.2</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>65</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>30.8</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>So, the weighted average is </h3><h3>62.93</h3><h3>×</h3><h3>69.2</h3><h3>%</h3><h3> </h3><h3>+</h3><h3> </h3><h3>64.93</h3><h3>×</h3><h3>30.8</h3><h3>%</h3><h3> </h3><h3>=</h3><h3> </h3><h3>63.55</h3><h3> </h3><h3>amu</h3><h3> .</h3><h3 /><h3>If we look at the Periodic Table, copper metal (a mixture of isotopes but </h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> and </h3><h3>65</h3><h3>C</h3><h3>u</h3><h3> predominate) has an approximate atomic mass of </h3><h3>63.55</h3><h3> </h3><h3>g</h3><h3>⋅</h3><h3>m</h3><h3>o</h3><h3>l</h3><h3>−</h3><h3>1</h3><h3> , so we know we are right.</h3>

4 0
3 years ago
Read 2 more answers
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