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zheka24 [161]
3 years ago
7

A test tube can be used to hold

Chemistry
1 answer:
Andrej [43]3 years ago
5 0
What grade is this? I forgot but if u tell me the grade i can prob remember
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How many grams of sodium nitrate. NaNO3, are soluble in 100
natka813 [3]

Answer:

Your correct answer is D. 40 g

Explanation:

Please mark brainliest!

4 0
3 years ago
Read 2 more answers
Zn + 2 HCl → ZnCl2 + H2
MrRissso [65]

Answer:

10.1g of H₂ are produced

Explanation:

To solve this question we need, first, to convert the mass of each reactant to moles and, using the chemical reaction, find limiting reactant. With limiting reactant we can find the moles of H2 and its mass:

<em>Moles Zn -Molar mass: 65.38g/mol-:</em>

307g * (1mol / 65.38g) = 4.696 moles

<em>Moles HCl -Molar mass: 36.46g/mol-:</em>

381g HCl * (1mol / 36.46g) = 10.45 moles

For a complete reaction of 10.45 moles of HCl are required:

10.45 moles HCl * (1mol Zn / 2mol HCl) = 5.22 moles Zn

As there are 4.696 moles of Zn, <em>Zn is the limiting reactant</em>

<em />

The moles of H₂ produced = Moles of Zn added = 4.696 moles. The mass is-Molar mass H₂ = 2.16g/mol-:

4.696 moles * (2.16g / mol) =

<h3>10.1g of H₂ are produced</h3>
7 0
3 years ago
Describe a Sankey Diagram and how it can be used.
jolli1 [7]

Explanation:

Sankey diagrams , which are typically used to visualize energy transfers between processes, are named after the Irishman Matthew H. P. R. Sankey, who used this type of diagram in a publication on energy efficiency of a steam engine in 1898.

Sankey diagrams are ideal for visually representing energy balances.

how to use

1.Overview. The Sankey diagram displays how quantities are distributed among items between two or more stages.

2.Add a Sankey diagram. Choose the Data Visualization or Re-Visualize option from the toolbar and select Sankey Diagram.

3.Change link color and width.

4.Change node color.

5.Change labels and tooltips.

4 0
3 years ago
How many moles of H2SO4 are needed to completely neutralize 0.0164 mol KOH
Vanyuwa [196]
0.0082 You have to equal out the amount of concentration with the unknown
8 0
4 years ago
Read 2 more answers
I’ve been stuck on these 5 questions!? Can you guys help?!
Varvara68 [4.7K]

Answer:

1. 0.224 moles of oxygen

3. 143.36 L oxygen gas

5. 0.059 atm

10. 5.14 atm

11. 307 K

Explanation:

1. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for n, which is n=(PV)/(RT)

P= 28.3 atm

V=0.193 L

R= 0.08206 L atm mol^-1 K^-1

T= 24.5+273.15= 297.65 K

Plugging the values in,

n=(28.3 atm x 0.193 L)/(0.08206 L atm mol^-1 K^-1 x 297.65 K)

n= 0.224 moles of oxygen

3. At STP, there are 22.4 L of gas for every mole of gas present. So 6.4 moles of oxygen would mean that there are:

6.4 mol x 22.4 L= 143.36 L oxygen gas

5. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is       P=(nRT)/V

n= 0.72 g converting to moles, divide by molar mass of oxygen gas:        0.72 g/32g= 0.0225 moles

V=9.3 L

R= 0.08206 L atm mol^-1 K^-1

T= 23.0+273.15= 296.15 K

Plugging the values in,

P=(0.0225 moles x 0.08206 L atm mol^-1 K^-1 x 296.15 K)/ 9.3 L

P= 0.059 atm

10. Ideal gas law again using the same equation as 5 above: You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is P=(nRT)/V

n= 0.108 mol

R=0.08206 L atm mol^-1 K^-1

T=20.0+273.15= 293.15 K

V= 0.505 L

Plugging the values in,

P=(0.108 mol x 0.08206 L atm mol^-1 K^-1 x 293.15 K)/0.505 L

P= 5.14 atm

11. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for Y, which is T=(PV)/(nR)

P= 0.988 atm

V= 1.20 L

n= 0.0470 mol

R=0.08206 L atm mol^-1 K^-1

Plugging the numbers in,

T=(0.988 atm x 1.20 L)/(0.0470 mol x 0.08206 L atm mol^-1 K^-1)

T= 307 K

4 0
3 years ago
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