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2 years ago

Which is an acceleration?

Chemistry

2 answers:

zhuklara [117]2 years ago

4 0

B hope that can help you

Oksana_A [137]2 years ago

3 0

(B) Because acceleration is change of speed over time.

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Alka-Seltzer is a solid mixture of two substances, sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7). The substances react

laiz [17]

1) List the reactants: sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇).

Reactants undergo change during a chemical reaction.

2) List the products: water (H₂O), carbon dioxide (CO₂) and sodium citrate (Na₃C₆H₅O₇).

Products are the substances formed from chemical reactions.

3) The balanced chemical equation:

3NaHCO₃ + H₃C₆H₅O₇ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇.

6 0

3 years ago

What is the name of the rock?<br> a. limestone<br> b. breccia<br> c. sandstone<br> d. conglomerate

Virty [35]

A is the best answer
Limestone

6 0

2 years ago

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Grace [21]

I relate to this so much

8 0

3 years ago

8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c

dangina [55]

Answer:

Partial pressure of $O_{2}$ in the gas was 733 torr and mass of $KClO_{3}$ in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$ )

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of $O_{2}$ = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that $O_{2}$ behaves ideally. Hence-

PV=nRT

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

So, $n=\frac{PV}{RT}$

= $\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}$

= 0.0259 moles

As 3 moles of $O_{2}$ are produced from 2 moles of $KClO_{3}$ therefore 0.0259 moles of $O_{2}$ are produced from $(\frac{2\times 0.0259}{3})$ moles or 0.0173 moles of $KClO_{3}$ .

Molar mass of $KClO_{3}$ = 122.55 g

So mass of $KClO_{3}$ in sample = $(0.0173\times 122.55)g$

= 2.12 g

7 0

2 years ago

2AlF3 + 3K2O → 6KF + Al2O3<br><br> How many grams of AlF3would it take to make 15.524 g of KF?

mr_godi [17]

<h3>Answer:</h3>

7.4797 g AlF₃

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

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

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0

2 years ago

Read 2 more answers