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2 years ago

What is occurring when reactants are mixed and heated and liquid collects in the sidearm of the apparatus

1 answer:

4 0

When reactants are mixed and heated and liquid collects in the sidearm of the apparatus, a chemical reaction is occurring.

<h3>What is a chemical reaction?</h3>

A chemical reaction is said to occur when reactants are combined to form products. We must note that the formation of products is evidenced by;

Appearance of a liquid
Appearance of a gas
Appearance of a solid

Hence, when reactants are mixed and heated and liquid collects in the sidearm of the apparatus, a chemical reaction is occurring.

Learn more about chemical reaction:brainly.com/question/22817140

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If i run away is that stealing from my parents?

victus00 [196]

Answer:

uh no I guess because your not your parents property your their child

4 0

3 years ago

Read 2 more answers

In the laboratory you are asked to make a 0.303 m cobalt(II) sulfate solution using 275 grams of water. How many grams of cobalt

valentinak56 [21]

Answer: 12.92g of CoSO4

Explanation:

Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol

Molarity of CoSO4 = 0.303mol/L

Mass conc. In g/L = Molarity x molar Mass

= 0.303x155 = 46.965g/L

275 grams of water = 0.275L of water

46.965g of CoSO4 dissolves in 1L

Therefore Xg of CoSO4 will dissolve in 0.275L i.e

Xg of CoSO4 = 46.965x0.275 = 12.92g

Therefore 12.92g of CoSO4 is needed

4 0

3 years ago

How many grams of H2O are needed to produce 150 g of Mg(OH)2? (Molar mass: H2O = 18.02 g/mol; Mg(OH)2 = 58.33 g/mol )

Nuetrik [128]

Answer:

92.6

Explanation:

6 mol x 18.02 g of H2o --> 3 mol x 58.33 g Mg(OH)2

108.12 g of h2o --> 174.99 of Mg(OH)2

g of H2O is 150 g of Mg(OH)2

150g x 108.12g / 174.99 =

92.67

7 0

3 years ago

When liquid heats up is it a chemical change

Mazyrski [523]

Yes because of the gas combination.

5 0

3 years ago

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0

2 years ago