Answer:
2.81 × 10⁶ mm³
2.81 × 10⁻³ m³
Explanation:
Step 1: Given data
Length (l): 250 mm
Width (w): 225 mm
Thickness (t): 50 mm
Step 2: Calculate the volume of the textbook
The book is a cuboid so we can find its volume (V) using the following expression.
V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³
Step 3: Convert the volume to cubic meters
We will use the relationship 1 m³ = 10⁹ mm³.
2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³
Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
I want to say A only... Hope I helped!!
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>
B; Seawater mixes with freshwater so the water has intermediate salinity
Explanation:
In an estuary, seawater mixes with freshwater so the water has intermediate salinity. Estuaries are usually located in transitional environments.
- Estuary is the wide part of a river where it nears the sea.
- This is called a transitional zone.
- Water from continental rivers usually fresh are brought in close contact with ocean water that is salty.
- The water here is said to be brackish as it is intermediate between salt and seawater.
- Organisms living in such terrain must be be well adapted to changing salinity.
Learn more:
salinity and density brainly.com/question/10491444
#learnwithBrainly