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nadezda [96]
2 years ago
9

Need help asap. Will vote brainly

Chemistry
1 answer:
andrew11 [14]2 years ago
8 0

Answer:

Answer

Explanation:

The second one is three, the last one is four, six is nine, fouris ten, eight is one, three is eight, seven is seven, five is five, three is six, one is two.

PLS Im only partly sure about a few but this is my answer.

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*URGENT*
DaniilM [7]

Answer:

d hope this helps u hhwjs GD jehehj

4 0
3 years ago
Read 2 more answers
Plete the following problems significant figures: 6 × 0.30<br>​
jeka57 [31]
6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.

0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.
3 0
3 years ago
Which statement describes how NO2- reacts in this equilibrium:
WITCHER [35]
The statement which describes how NO2- reacts in this equilibrium:
<span>H2SO3(aq) + NO2-(aq) HSO3-(aq) + HNO2(aq
is the second option - </span><span>B. as a Brønsted-Lowry base by accepting a proton.
</span>This is because bases take proton H+ in order to become HNO2. 
6 0
3 years ago
Read 2 more answers
A solution is made by dissolving 10.7 g of magnesium sulfate, MgSO4, in enough water to make exactly 100 mL of solution. Calcula
adell [148]

Answer:

[MgSO₄] = 890 mM/L

Explanation:

In order to determine molarity we need to determine the moles of solute that are in 1L of solution.

Solute: MgSO₄ (10.7 g)

Solvent: water

Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L

We convert the solute's mass  to moles → 10.7 g / 120.36 g/mol = 0.089 moles

Molarity (mol/L) → 0.089 mol/0.1L = 0.89 M

In order to calculate M to mM/L, we make this conversion:

0.89 mol . 1000 mmoles/ 1 mol = 890 mmoles

4 0
3 years ago
Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is m
12345 [234]

<u>Answer:</u> The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For butane:</u>

Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol

The chemical equation for the reaction of butane and oxygen gas follows:

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with = \frac{13}{2}\times 0.224=1.456mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce = \frac{10}{2}\times 0.224=1.12moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams

8 0
4 years ago
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