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KiRa [710]
3 years ago
6

For a particular experiment, Maria must heat

Chemistry
2 answers:
mariarad [96]3 years ago
5 0

Answer:

Heat Proof Gloves

Explanation:

Heat-resistant gloves are work gloves designed to protect the hands from burns or other injuries that can result from coming into contact with extremely hot objects

Ilya [14]3 years ago
5 0

Answer:

goggles

a lab coat or apron

heatproof gloves

Explanation:

just got it wrong because the first answer was incorrect

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2 Cu + Cl2 ----> 2 Cuci
s344n2d4d5 [400]

Answer:

320 g.

Explanation:

Hello!

In this case, according to the balanced chemical reaction, we can compute the grams of copper I chloride produced by each reactant, as shown below:

m_{CuCl}^{by\ Cu}=3.23molCu*\frac{2molCuCl}{2molCu}*\frac{99.0gCuCl}{1molCuCl}  =320gCuCl\\\\m_{CuCl}^{by\ Cl_2}=1.64molCl_2*\frac{2molCuCl}{1molCl_2}*\frac{99.0gCuCl}{1molCuCl}  =325gCuCl

Thus, since copper produces the fewest grams of CuCl, we infer it is the limiting reactant, therefore the correct mass of copper I chloride is 320 g.

Best regards!

4 0
3 years ago
How many atoms are there in 1 g of argon?
amm1812

Answer:

1.5057×10^22 atom

Explanation:

As we

1 mole of argon = 40 g of argon

i.e 40 g of argon = 1 mole of argon

1 g of argon = 1/40 mole of argon

1 mole of argon = 6.023×10^23 atom of argon

1/40 mole if argon = 1/40 ×6.023×10^23

= 1.5057×10^22

7 0
3 years ago
10. A sample of an unknown composition was tested in a laboratory. The sample could not be broken down by
damaskus [11]

Answer:

it would be an element because its an element

Explanation:

8 0
3 years ago
Read 2 more answers
You need 270 ml of a 65% alcohol solution. on hand, you have a 90% alcohol mixture. how much of the 90% alcohol mixture and pure
olganol [36]

You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.

<em>Step 1.</em> Calculate the volume of 90 % alcohol needed

You can use the dilution formula

<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2

where

<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions

<em>C</em>1 and <em>C</em>2 are the concentrations

You can solve the above formula to get

<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2

<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %

V2 = ?; _____<em>C</em>2 = 90 %

∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL

You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA

<em>Step 2</em>. Calculate the amount of water to add.

Volume of water = 270 mL – 195 mL = 75 mL

8 0
3 years ago
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
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