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leva [86]
3 years ago
10

What animals are in the Arctic Tundra?

Chemistry
2 answers:
ehidna [41]3 years ago
6 0

Answer:

Polar Bears

Arctic Foxes

Seals

Penguins

They all live in the cold, so they all must live in the Arctic Tundra ;)

loris [4]3 years ago
6 0

Answer:

Animals such as polar bears, arctic wolves, and arctic foxes live in the Arctic Tundra.

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At room temperature, compounds that are held together by covalent bonds tend to exist in what state?
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at room temperature substances hawks together by covalent bonds are typically in liquid form

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Wood, Iron, Glass, Paper, Nickel Which substances show magnetic property?
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The comet cools down and begins to glow brightly as a result of this decreased temperature.
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Calcium carbonate can be decomposed by heating: CaCO3 - CaO + CO2
Galina-37 [17]

Answer: 13.888g

Explanation:

CaCO3 —> CaO + CO2

From the equation above, 1mole of CaCO3 produces 1mole of CaO. Since the answer is to be expressed in grams, let us covert this mole to grams. This is illustrated below:

MM of CaCO3 = 40 +12 + (16x3) = 40 + 12 + 48 = 100g/mol

MM of CaO = 40 + 16 = 56g/mol

From the equation,

100g of CaCO3 produced 56g CaO

Therefore, 24.8g of CaCO3 will produce Xg of CaO i.e

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Therefore, the theoretical yield of CaO is 13.888g

3 0
3 years ago
The rate law for the reaction 2A + B →C is –rA= kACA2CBwithkA= 25 (L/mol)2sec. What arekBandkC?
givi [52]

Explanation:

The given reaction equation is as follows.

             2A + B \rightarrow C

So, rate constants for different reactants and products written as follows.

             \frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}

As per the reaction equation, the stoichiometric coefficients of reactants and products are as follows.

         A = -2

         B = -1

         C = 1

Therefore,

      \frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}

      \frac{-k_{A}}{-2} = \frac{-k_{B}}{-1} = \frac{k_{C}}{1}    

             \frac{-k_{A}}{-2} = k_{B} = k_{C}

Hence,          k_{B} = k_{C} = \frac{25}{2} (L/mol)^{2}

                                          = 12.5 (L/mol)^{2}

Thus, we can conclude that k_{B} and k_{C} are 12.5 (L/mol)^{2}.

5 0
3 years ago
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