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quester [9]
2 years ago
6

What type of mirror should be used for this application? Explain why this mirror is the right choice for a telescope.

Chemistry
1 answer:
Artyom0805 [142]2 years ago
7 0

Answer:

use a concaved miror

Explanation:

better and more clear

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How many Mol are in 4.000 grams of MgCl2 ?
grigory [225]

Answer:

The number of mol is: 0, 042 mol in 4 grams of MgCl2

Explanation:

We calculate the weight of 1 mol of MgCl2:

Weight 1mol of MgCl2= weight Mg + (weight Cl)x  2=

24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2

95, 3 grams------1 mol MgCl2

4 grams      -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2

4 0
3 years ago
The difference between the boiling point of a pure solvent and the boiling point of a solution of a nonelectrolyte in the same s
Sati [7]
Answer: A) Boiling-Point Elevation
6 0
2 years ago
15.0 g of Fe and 25.0 g of sand are added to 250.0 g of water. a. Determine the percent mass of Fe, sand, and water in the mixtu
lana [24]

Answer:

A. percentage mass of iron = 5.17%

percentage mass of sand  = 8.62%

percentage mass of water = 86.205%

B. (Iron + sand + water) -------> ( iron + sand) ------> sand

C. The step of separation of iron and sand

Explanation:

A. Percentage mass of the mixtures:

Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g

percentage mass of iron = 15/290 * 100% = 5.17%

percentage mass of sand = 25/290 * 100% = 8.62%

percentage mass of water = 250/290 * 100% = 86.205%

B. Flow chart of separation procedure

(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand

C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.

5 0
2 years ago
trial 2, 2.68 g/cm/3 trial 3, 2.84g/cm/3. aluminum has a density of 2.70g/cm/3 calcute the percent error for each trial.
serg [7]

Answer:

trial 2: 0.74

trial 3: 5.19

I think but the equation for sloving percent error is:

(true value - determined value)/true value * 100

Explanation:

6 0
3 years ago
If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many
vladimir2022 [97]
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

<span>Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
--------------------- =      ---------------------------
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
----------------------- =  ----------------------------
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.
</span>


8 0
3 years ago
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