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3 years ago

14

Kelly has some sand in a bucket, but she thinks that there may be some salt mixed in with her sand. How can she determine if the

re is salt in the sand? A. If she leaves the bucket alone the salt will rise to the top. B. If she puts water in the bucket, the sand will dissolve. C. If she puts water in the bucket, the salt will dissolve. D. If she evaporates the sand the salt will be left behind.

2 answers:

Nana76 [90]3 years ago

7 0

C)............. Hope it helps, Have a nice day:)

Gre4nikov [31]3 years ago

6 0

D i think i hope this helps

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It has 4 valence electrons

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What is produced during the replacement reaction of ba(no3)2 and na2so4? 2bana 2no3so4 2nano3 baso4 nano3 baso4 bana2 (no3)2so4

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2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Procedures involved:

The cations or anions may transfer positions in this twofold replacement/displacement reaction, which results in AB + CD AD + CB. In such a reaction, water, an insoluble gas, or an insoluble solid must be one of the byproducts (precipitate). The reaction in question has the following molecular equation:

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When two atoms or groups of atoms swap positions, a double displacement reaction occurs, creating new compounds. Typically, aqueous solutions are where it happens.

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2 years ago

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3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

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Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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3 years ago

How many moles are represented by 3.01 x10^24 oxygen atoms?

asambeis [7]

<h3>Answer:</h3>

5.00 mol O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.<u> </u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.01 × 10²⁴ atoms O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.01 \cdot 10^{24} \ atoms \ O_2(\frac{1 \ mol \ O_2}{6.022 \cdot 10^{23} \ atoms \ O_2})$
Multiply/Divide: $\displaystyle 4.99834 \ mol \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

4.99834 mol O₂ ≈ 5.00 mol O₂

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3 years ago