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2 years ago

A substance that can be separated into two or more substances only by a chemical change is

1 answer:

5 0

A substance that can be separated into two or more substances only by a chemical change is known as a heterogeneous mixture

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What does this picture represent?

WARRIOR [948]

the answer is d electronic balance

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2 years ago

What part of an atom is involved in a chemical reaction?

Katyanochek1 [597]

Answer:

The answer is supposed to be "Electron cloud" or "Electon".

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3 years ago

Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?

Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of CaCO₃ with HCl.

CaCO₃ : HCl

1 : 2

0.087 : 2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0

2 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

2 years ago

How many grams are in 0.35 moles of C2H4? please show dimensional analysis

fredd [130]

Answer:

9.8g

Explanation:

Using periodic table find molar mass of C and H:

C=12.01g/mol

H=1.008g/mol

Molar mass of C2H4=(12.01)2+(1.008)4=28.03g/mol

Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:

28.03 x 0.35=9.8grams

5 0

2 years ago