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Reika [66]
3 years ago
9

8.45 cm - 3.55 cm = ?

Chemistry
2 answers:
lyudmila [28]3 years ago
6 0

Answer:

4.9cm or 4.90cm

Explanation:

Use column method and align the numbers. Make sure you add the decimal point before you write the answer, otherwise the value would be wrong.

mafiozo [28]3 years ago
5 0

Answer:

4.90

Explanation:

8.45.

-3.55

minus those two and you'll get the sum!

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Isotopes percent abundance of antimony
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Antimony has two naturally occurring isotopes. Their abundance is given in the pic attached below
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3 years ago
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe
polet [3.4K]
First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
                                                             = 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
                                                                  = 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
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4 0
3 years ago
A student calculates the volume of a graduated cylinder to be 43.26 ml. The actual volume is 42.32 ml. What is the percent error
Tpy6a [65]

Answer:

2.2%

Explanation:

Percentage error,

You apply the formula,

[(Estimated value - Actual value)/Actual value] × 100%

; [(43.26 - 42.32)/42.32] × 100

; (0.94/42.32) × 100

; 0.022 × 100

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After the death of living material, the ratio of carbon-12 to carbon-14 isotopes in the material;
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3 years ago
You are given a 1.55 g mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 20 mL of water and add an e
irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.

<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>

0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

<em>Moles CaCl₂:</em>

3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>

1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture

That means mass percent of CaCl₂ is:

0.207g CaCl₂ / 1.55g * 100 =

<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>
8 0
3 years ago
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