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pav-90 [236]
3 years ago
11

Will ice melt faster with a lot of salt added or little salt added?

Chemistry
1 answer:
STatiana [176]3 years ago
5 0

Answer:

alot

Explanation:

because salt contains stuff

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The bond that describes an interaciton between two orbitals that is not symmetrical about a line between the two atoms nuclei is
inna [77]

Answer:

Pi bond

Explanation:

Pi bond is the type of covalent bond which results from formation of molecular orbital by the side-to-side overlap of the atomic orbitals along the plane perpendicular to the line which connects nuclei of the two atoms in which the bond is formed. It is denoted by symbol π.

The bond is not symmetrical to the internuclear axis and on rotation the axis, the bond breaks.

3 0
3 years ago
A 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O. Find the mass of sulfur in a sample of CaSO4 with a m
andrey2020 [161]

The mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

<h3>How to calculate mass of an element in a compound?</h3>

According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.

This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g

Next, we calculate the percent ratio of each element in the compound; CaSO4.

  • Ca = 3.06g/10.4g × 100 = 29.42%
  • S = 2.45g/10.4g × 100 = 23.56%
  • O = 4.89g/10.4g × 100 = 47.02%

According to this question, a sample of CaSO4 with a mass of 65.8 g is given. The mass of each element in this compound is as follows:

  • Ca = 29.42/100 × 65.8g = 19.36g
  • S = 23.56/100 × 65.8g = 15.50g
  • O = 47.02/100 × 65.8g = 30.94g

Therefore, the mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

Learn more about mass at: brainly.com/question/13672279

#SPJ1

5 0
2 years ago
CAN SOMEOME GIVE ME AN EXAMPLE OF
n200080 [17]

Answer:

- A ball does not move unlesss you kick it and it rolls aways

- a car stays still and is not in motion, unless you start the engine

- a pencil stays still on a table, it moves when you pick it up off the table

Explanation:

7 0
3 years ago
Write the formula equation for the reaction between hydroiodic acid and beryllium hydroxide
Aleksandr-060686 [28]
 The  formula  equation for  the reaction  between  hydroiodic  acid  and  beryllium  hydroxide      is as below

Be(OH)2  +  2 HI   =  BeI2 +   2H2O


Formula  of  hydroiodic is  =  HI
formula   of  beryllium hydroxide = Be(OH)2


1 mole  of  Be(OH)2  react  with   2  mole  of HI to form  1 mole  of BeI2  and   2   mole  of H2O
8 0
4 years ago
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ
NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂   (1)        

b) The precipitate formed is Ni(OH)₂  

 

c) The limiting reactant is:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

5 0
4 years ago
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