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iren2701 [21]
3 years ago
13

Indicate the answer choice that best completes the statement or answers the question.

Chemistry
2 answers:
8090 [49]3 years ago
5 0

Answer: A

Explanation:

Jlenok [28]3 years ago
5 0

Answer:

The answer is question 100

Explanation:

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Can someone help me with 3
lozanna [386]
Solidification i believe
5 0
3 years ago
Find the mass of 0.235 moles of SiO2? (NOTE: use correct significant figures!)
marta [7]

Answer: 14.1g

Explanation:

Given that,

number of moles of SiO2 = 0.235 moles

Mass in grams = Z (let unknown value be Z)

Molar mass of SiO2 = ?

To get the molar mass of SiO2, use the atomic mass

Silicon = 28g;

Oxygen = 16g

i.e Molar mass of SiO2 = 28g + (16g x 2)

= 28g + 32g

= 60g/mol

Now, apply the formula

Number of moles = Mass / molar mass

0.235 moles = Z / 60g/mol

Z = 0.235 moles x 60g/mol

Z = 14.1 g

Thus, the mass of SiO2 is 14.1 grams.

8 0
3 years ago
2CO + O2 --> 2CO2
GREYUIT [131]

Answer:

10 L of CO₂.

Explanation:

The balanced equation for the reaction is given below:

2CO + O₂ —> 2CO₂

From the balanced equation above,

2 L of CO reacted to produce 2 L of CO₂.

Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:

From the balanced equation above,

2 L of CO reacted to produce 2 L of CO₂.

Therefore, 10 L of CO will also react to produce 10 L of CO₂.

Thus, 10 L of CO₂ were obtained from the reaction.

3 0
3 years ago
PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things
harkovskaia [24]

Answer:

organic compounds contained in all living things

Explanation:

3 0
3 years ago
Read 2 more answers
How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of
alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0
3 years ago
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