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amm1812
3 years ago
8

What is the mass of a light bulb

Chemistry
1 answer:
katen-ka-za [31]3 years ago
3 0
The mass of a lightbulb is 6.2 ounces.
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Why air is classified as a mixture​
Vika [28.1K]

Answer:

See explanation

Explanation:

Air contains a mixture of several molecules and compounds such as oxygen and carbon dioxide.

5 0
3 years ago
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0
2 years ago
What is the gcf of 16 and 72? <br> a. 2 <br> b. 4 <br> c. 16 <br> d. 8
nadya68 [22]
Your answer is D. 8

16 = 2^4
72 = 2^3*3^2

So you'll choose 2^3 = 8
7 0
4 years ago
Read 2 more answers
Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.
Anastaziya [24]

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

Step 3: Calculate the pH

pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15

7 0
3 years ago
An analytical chemist weighs out 0.055g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with
Katarina [22]

Answer:

Mass of the unknown acid is 4.0g

Explanation:

The determine the molar mass of the unknown acid, the steps below can be followed

Firstly, determine the concentration of the acid, the formula below can be used;

ConcA × Va/ConcB × Vb = Na/Nb

Where ConcA is the concentration of the unknown acid

ConcB is the concentration of the NaOH base

Va is the volume of acid and Vb is the volume of base

Since, the titration was said to have reached an equivalent point, it means the number of moles of the acid (Na) was equal to the number of moles of the base (Nb) and thus both will be assumed to be 1

Thus

ConcA × 250/0.13 × 6.6 = 1/1

ConcA = 0.13 × 6.6/250

ConcA = 0.003432M

Then, determine the actual number of moles (n) of the unknown acid used,

ConcA = no of moles of acid/volume of acid (in dm³ or L)

To convert mL to L, we divide by 1000

Hence, 250ml = 0.25L

0.003432 = n/0.25

n = 0.003432 × 0.25

n = 0.01373 moles

To determine the molar mass;

n = mass/molar mass

The mass was given in the question to be 0.055g

Thus

0.01373 = 0.055/molar mass

molar mass = 0.055/0.01373

molar mass = 4.0g

5 0
3 years ago
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