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Kryger [21]
3 years ago
8

What is simple chemical test could you carry out to see if carbon dioxide has been produced

Chemistry
2 answers:
larisa86 [58]3 years ago
7 0

Carbon Dioxide

A positive test will result in the lime water turning milky. Lime water turns milky as the Calcium hydroxide (chemical name for limewater) reacts with carbon dioxide to form Calcium Carbonate which is insoluble in water and thus forms a milky white precipitate.

levacccp [35]3 years ago
5 0
The answer is lime water turning milky ;)
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Gather and respond to information
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2 years ago
A single covalent bond is made up of​
SIZIF [17.4K]

Answer:

In chemistry, a single bond is a chemical bond between two atoms involving two valence electrons. That is, the atoms share one pair of electrons where the bond forms. Therefore, a single bond is a type of covalent bond.

Explanation:

(copied from Google)

4 0
3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

4 0
3 years ago
را
Aloiza [94]

Answer:

A.It is the same for every sample of a single substance.

Explanation:

7 0
3 years ago
A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
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