How many grams of CuSO4 are there in 100.0g of hydrate?

Answers

Answer 1

Answer: Usually copper(ii) sulfate has a hydrate form with 5 moles of water or copper (II) Sulfate pentahydrate or CuSO4.5H2O. To answer this, 100g of the hydrate is multiplied by the reciprocal of its molar mass then muliplied by the molar ratio of CuSO4 to CuSO4.5H2O. Then finally

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All radioactive nuclides undergo what

Answers

They undergo nuclear fission.

Find the volume of a cube of zinc with the following dimensions 3.000 cm 3.1 cm 2.99 cm

Answers

i think it would be 27cm but im not sure, its been a while since ive done this -A

Final answer:

The volume of the cube of zinc is 27.687 cubic cm. Multiply the given dimensions together to find the volume of the 'cube' of zinc.

Explanation:

The question asks for the volume of a hypothetical cube of zinc with side lengths 3.000 cm, 3.1 cm, and 2.99 cm. While a cube should have all sides of equal length, we will proceed with the given dimensions. The volume of a rectangular parallelepiped (in our case, a cube-like shape) can be found by multiplying its length, width, and height together. So, the volume of this zinc 'cube' would be calculated as follows: 3.000 cm x 3.1 cm x 2.99 cm = 27.687 cubic cm.

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Which will cause a decrease in gas pressure in a closed containerA) Reducing the volume
B) Adding more gas
C) Lowering the temperature

Answers

By lowering the temps, there is a decrease in the pressure inside. so C.

Answer: C, Lowering the temperature

Explanation: Gradpoint

Calculate the pH of an aqueous solution of 16 mL of 0.00046 M HClO4(aq) after dilution to 42 mL.

Answers

1000 $cm^(3)$ of HClO_{4} [/tex] contain 0.00046mol
16 $cm^(3)$ of HClO_{4} [/tex] contain x
x = $(16cm ^(3) * 0.00046mol)/(1000cm x^(3) )$
x = 0.00000736mol

$M_(1) V_(1) = M_(2) V_(2)$
(0.00000736mol)(16 $cm^(3)$ )= $(x) (42cm^(2))$
$x= ((0.00000736mol) * (16cm^(3)) )/(42cm^(3) )$
= 0.000002803mol

pH= -log [ $H^(+)$ ]
= -log (0.000002803cm $^(3)$
= 5.55

What is the maximum mass of methane (CH4) that can be burned if only 1.0 g of oxygen is available and how do you figure it out?

Answers

Make the reaction: $CH_4+2O_2=CO_2+2H_2O$
For 2*32=64 grams of oxygen, you can burn 2*16=32 grams of methane (that is, half).
So, our answer is .5g.
Other intermediate calculations are mollecular masses:
$\mu_(CH_4)=A_C+4A_H=12+4=16 \n mu_(O_2)=2A_O=2*16=32$

Answer:

0.28 g of $CH_4$

Explanation:

You need a balanced equation first. $CH_4 + 2O_2$ ⇒ $CO_2 + 2H_2O$

You need to find the number of moles of oxygen, which is mass divided by the Mr: 1 ÷ 32 = 1/32

Then you find the moles of methane using the mole ration oxygen to methane 2:1. 1/32 ÷ 2 = 1/64

Then you find the mass by multiplying the Mr with the number of moles:

1/64 × 18 = 0.28125 g = 0.28 g

The sum of pH and pOH for any aqueous solution equals

Answers

pH is a notion that expresses the concentration of hydrogen in a solution while pOH expresses the concentration of hydroxide in a solution. The sum of pH and pOH in a solution is 14.

Answer: 14

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.

$pH=-\log [H^+]$

$pOH=-\log[OH^-]$

$pH+pOH=14$

Thus the sum of pH and pOH is 14.

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