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2 years ago

What is the m∠L? 45° 54° 40° 52°

Chemistry

1 answer:

Nataly [62]2 years ago

4 0

Answer:

m<L = 45

Explanation:

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Which group of factors are most important in determining the composition of ocean water? a.)temperature, salinity, and density b

Elis [28]

Temperature, salinity, and density are the group of factors are most important in determining the composition of ocean water.

a.)temperature, salinity, and density

<u>Explanation:</u>

The three fundamental factors that help in determining the composition of ocean water are temperature, salinity, and density. Temperature, saltiness, salinity, and density influence the thickness of seawater.

Enormous water masses of various densities are significant in the layering of the sea water (increasingly thick water sinks). As temperature builds water turns out to be less thick. As saltiness builds water gets denser. The temperature helps in deciding the pace of vanishing of the ocean.

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A. Thermal energy good job

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Identify the products formed in this Brønsted-Lowry reaction.<br><br> HPO2−4+NO−2↽−−⇀acid+base

klemol [59]

Answer: The products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Explanation:

According to Bronsted-Lowry, acids are the species which donate hydrogen ions to another specie in a chemical reaction.

Bases are the species which accept a hydrogen ion upon chemical reaction.

For example, $HPO^{2-}_{4} + NO^{-}_{2} \rightleftharpoons HNO_{2} + PO^{2-}_{4}$

Here, the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Thus, we can conclude that the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

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3 years ago

What is the percent by mass of carbon in acetone, c 3 h 6 o?

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In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

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H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C= $( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%$
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%

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Please help!! Why is the following Electron Configuration incorrect for Aluminium?

Margarita [4]

Answer:

3s^1 would be 3s^2

Explanation:

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