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hram777 [196]
3 years ago
6

Suppose we want to charge a flask with 1.6 g of sugar. We put the empty flask on a balance and it is determined to weigh 330 g.

Enter the weight we would expect to see on the balance when we're done adding the sugar.
Chemistry
2 answers:
Olin [163]3 years ago
7 0

Answer:

We expect 331.6 grams on the balance

Explanation:

Step 1: Data given

Mass of the empty flask = 330 grams

We want a mass of 1.6 grams of sugar

Step 3: Calculate the total mass

Total mass = mass of empty flask + mass of sugar

Total mass = 330 grams + 1.6 grams sugar

Total mass = 331.6 grams

The total mass of the flask with the sugar should be 331.6 grams

When we have this on the balance we have 1.6 grams of sugar in the flask.

RSB [31]3 years ago
5 0

Answer:

total weight to expect = 331.60 grams

Explanation:

The flask wanted to be charged with 1.6 grams of sugar. The empty flask was weighed on a balance and it was determined to weigh 330 grams . The weight we would expect when we add the sugar can be computed as follows:

weight of the empty flask = 330 grams

weight of the sugar = 1.6 grams

The weight of the flask after adding the sugar will be the sum of the weight of the sugar and the empty falsk

Therefore,

Total weight to expect = 330 grams + 1.6 grams

total weight to expect = 331.60 grams

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La etiqueta de una bebida transparente con un ligero toque anaranjado indica que contiene aga con un 15 porciento de zumo de nar
Brilliant_brown [7]

<u>Respuesta:</u>

16 gramos de vitamina C y 8 gramos de colorantes.

Explanation:

Debemos multiplicar el porcentaje de cada componente por la cantidad total (100 %) de zumo.

800 g de zumo = 100%

- Tenemos un 2 porciento (2 en 100 o 2/100) de vitamina C. Por lo tanto, la cantidad en gramos de vitamina C es:

2/100 x 800 g = 16 g

- Tenemos un 1 porciento de colorantes en el zumo (1/100). Por lo tanto, la cantidad en gramos de colorantes es:

1/100 x 800 g = 8 g

6 0
3 years ago
Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3
maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

<em>HF = 1.62g</em>

<em>H₂O = 516g</em>

<em>F⁻ = 0.163g</em>

<em>H₃O⁺ = 0.110g</em>

<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

<h3>Kc = 1.09x10⁻⁴</h3>
7 0
3 years ago
I need help on this question
uranmaximum [27]

Answer:

When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.

Explanation:

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6 0
3 years ago
Read 2 more answers
Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
3 years ago
HCI + NaOH --&gt; NaCl + H 20
Vikki [24]

Answer:

option D is correct

Explanation:

no of moles in 3 grams of HCL=3/36=0.08

if 1 mole of HCL require 1 mole of NaOH then 0.08 moles required 0.08 moles of NaOH

mass of 0.08 moles of NaOH=moles*molar mass=0.08*40=3.2 grams

so 3 grams are required in the reaction

6 0
2 years ago
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