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seropon [69]
3 years ago
8

Litmus paper is made from water-soluble dyes which are extracted from lichens. This paper is used as an acid-base indicator. Whi

ch of these common household substances would turn blue litmus paper red? A) ammonia Eliminate B) vinegar C) baking soda D) distilled water
Chemistry
2 answers:
yarga [219]3 years ago
4 0

Your answer is vinegar C)

Dovator [93]3 years ago
4 0

Blue litmus paper turns red in the presence of an acidic solution/compound. An acid is any substance with a pH level less than 7.

Ammonia is a moderately basic solution. It has a pH greater than 7. Therefore, it will not turn blue litmus paper red.

Baking soda is also a basic substance with a pH greater than 7. It also will not turn blue litmus paper red.

Distilled water is a neutral solution. It has a pH of 7, and it will not react with the litmus paper to change its color.

That leaves us with vinegar. Vinegar is an acid with a pH of less than 7. It will turn blue litmus paper red.

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What mass of nitrogen dioxide, NO2, would be produced by the decomposition of 75.0 g of copper (II) nitrate?
astraxan [27]
Mole number of 75 g is 0.4 mol. NO2 is 0.8 mol and is 36.8 g.
6 0
3 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

4 0
3 years ago
How far would Sara get if she ran for 87 minutes at a speed of .125 miles per minute?
Elza [17]

Answer:

10.875 miles

So 87 minutes

.125 miles per minute

So just multiply int his case

87*.125

You’d get 10.875

7 0
3 years ago
Balance the reaction which forms ions. Choose "blank" if no coefficient other than 1 is needed. Pb(N03)2(aq) --&gt; ___ Pb+2 + _
aleksklad [387]
Pb(NO3)2 -> Pb+2 + 2 NO3-
8 0
4 years ago
Read 2 more answers
If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac
mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

<u />

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) +  6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0
4 years ago
Read 2 more answers
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