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3 years ago

What does the symbol "kg" represent in the metric system?

Chemistry

2 answers:

gizmo_the_mogwai [7]3 years ago

5 0

Answer: A

Explanation:

Ive done this one before

agasfer [191]3 years ago

3 0

$\huge\mathfrak\red {Answer:}$

It represents

<h2>A) 1,000 grams</h2>

(1 kilogram = 1000 grams)

$\mathfrak\purple {Hope\: this\: helps\: you...}$

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The pressure of 5.00 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant tempe

Vsevolod [243]

(P1)(V1)=(P2)(V2)
(1.50)(5.00)=(1240/760)(V2)
(7.5)/(1240/760)=V2
V2=4.596774194 L

5 0

3 years ago

Read 2 more answers

Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if

mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

<em>initial volume of gas in the container, V₁ = 2.65 L</em>
<em>initial number of moles of gas, n₁ = 0.12 mol</em>
<em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

$n_t = 0.12 + 0.182 = 0.302 \ mol$

The final volume of gas in the container is calculated as follows;

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L$

Thus, the final volume of the methane gas in the container is 6.67 L.

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5 0

3 years ago

How do you determine the volume of a regular object if you don't know the density?

Ann [662]

Answer:

mol times or devided by molar volume

4 0

3 years ago

Ice floats in water because it _____. is less dense than water contains more oxygen than water has a higher pH than water has mo

Tpy6a [65]

Answer:

it is less dense than water

Explanation:

odyssey ware

3 0

3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s) + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0

3 years ago