What’s the question sweetheart...?...There’s no question or picture to answer
Answer:
The final temperature of the water will be 328.81 K .
Explanation:
Using the equation, <em>q = mcΔT</em>
here, <em>q = energy</em>
<em>m= mass</em>
<em>c= specific heat capacity </em>
<em>ΔT= change in temperature</em>
<em>Mass of water = 1kg (1000 g ) per liter</em>
<em>∴ 6.2 Liter of water = 6200 g</em>
<em>c of water ≈ 4.18 J /g/K</em>
<em>Now, </em>
<em>980000 = 6200*4.18*ΔT</em>
<em>ΔT = 37.81 K</em>
<em>∴ final temperature of the water = 291 + 37.81 = 328.81 K</em>
In order to determine if the ion is positively charged or negatively charged
Answer:
839 g
Explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 81.97 209.94
NaAlO₂ + … ⟶ Na₃AlF₆ + …
Mass/g: 2150
1. Calculate the <em>moles of Na₃AlF₆
</em>
Moles of Na₃AlF₆ = 2150 × 1/209.94 Do the operation
Moles of Na₃AlF₆ = 10.24 mol Na₃AlF₆
2. Calculate the <em>moles of NaAlO</em>₂
The molar ratio is 1 mol NaAlO₂:1 mol Na₃AlF₆
Moles of NaAlO₂ = 10.24 × 1/1 = 10.24 mol NaAlO₂
3. Calculate the <em>mass of NaAlO₂
</em>
Mass of NaAlO₂ = 10.24 × 81.97 Do the multiplication
Mass of NaAlO₂ = 839 g
