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Ierofanga [76]
3 years ago
5

How many moles of SnF₂ will be produced along with 48 grams of H₂?

Chemistry
1 answer:
Cloud [144]3 years ago
4 0

Answer:

...................................................................................l..........................................

Explanation:

......................................................................................................................................................................................................................................................

You might be interested in
24.08 x 10^23 atoms of Boron (B) is equal to how many moles of Boron?
padilas [110]
1 mole ----------- 6.02 x 10²³ atoms
? mole ---------- 24.08 x 10²³ atoms

moles B = ( 24.08 x 10²³) x 1 / 6.02 x 10²³

moles B = 24.08 x 10²³ / 6.02 x 10²³

= 4 moles

Answer B

hope this helps!
5 0
3 years ago
Read 2 more answers
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the
densk [106]

Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

  • Q sln = mCΔT

∴ m sln = m Ba(OH)2 + m HCl

∴ molar mass Ba(OH)2 = 171.34 g/mol

∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol

⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g

∴ molar mass HCl = 36.46 g/mol

∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol

⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g

⇒ m sln = 3.084 g + 1.313 g = 4.3966 g

specific heat (C):

∴ C sln = C H2O = 4.18 J/g°C

∴ ΔT = 26.83°C - 22.74°C = 4.09°C

heat absorbed (Q):

⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)

⇒ Q sln = 75.165 J

8 0
3 years ago
Minerals can be easily identified by their?
kotegsom [21]
Im pretty sure it would be d.
7 0
3 years ago
Read 2 more answers
A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa
Phoenix [80]

Answer:

V_{2} = 8.92 L

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

P_{1} V_{1} =nRT

320Kpa*0.003m^{3} =nRT

1000L = 1m^{3}

So

nRT= 0.96

Now, as for the final condition:

P_{2}V_{2}=nRT

P_{2} V_{2} =0.96

clearingV_{2}

V_{2} =\frac{0.96}{P_{2} }

V_{2} =0.00872m_{3}

V_{2} = 8.92 L

6 0
3 years ago
Which of the following best describes the formation of plasma?
Sav [38]
I say, D is the answer

5 0
3 years ago
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