Answer:
5.17.
Explanation:
<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
</em>
[OH⁻] = 1.5 x 10⁻⁹ M.
∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5 x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.
∵ pH = - log[H₃O⁺]
<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17.
</em>
All organic substances contain carbon atoms.
Answer:
thick, insulating fur
Explanation:
If an animal lives in a freezing climate, it makes sense logically that the animal would adapt and develop a layer of thick fur to keep its body insulated and maintain homeostasis.
Hope this helped and please consider a Brainliest! :)
I think the reaction that is occurring is CuSO₄(aq)+Na₂CO₃(aq)⇒CuCO₃(s)+Na₂SO₄(aq). Water in the hydrate will just become part of water in solution so it does not really matter for this question.
SInce we now know the balanced chemical equation, you can use stoichiometry to find the amount of reactants needed to produce 5g CuCO₃.
The first step to any stoichiometry question is to convert the mass given into moles. To do this you have to divide the mass by the molar mass of the compound the mass is referring to. The molar mass of CuCO₃ is 123.5g/mol so you have to divide 5g by 123.5g/mol to get 0.04047 moles of CuCO₃.
The next step in this qustoin is to find the number of moles of each reactant using the moles of product we found in the first step. To do this, we need go back and look at the equation to find the molar ratios which is shown through the coefficients. Since all of the coeficients in the chemical equation is 1, we know the molar ratios are all `1 to 1 and therefore the number of moels of each reactant is equal to the number of moles of the product found in step one. This means that you started off with 0.04047 moles of both CuSO₄ and Na₂CO₃.
The final step is to multiply the number moles of each reactant by its molar mass. the molar mass of CuSO₄ is 159.6g/mol and the molar mass of Na₂CO₃ is 106g/mol. When you multiply 0.04047mol by 159.6g/mol you get 6.4g and when you multiply 0.04047mol by 106g/mol you get 4.29g.
Therefore you started with 6.4g of CuSO₄ and 4.29g of Na₂CO₃.
I hope this helps.
