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3 years ago

Is mixing butter, granulated sugar, and brown sugar until it's creamy a physical or chemical change and why

Chemistry

2 answers:

Nadusha1986 [10]3 years ago

8 0

Answer:

Physical

Explanation: Because it says "Until creamy" meaning you're changing the look, not the way it reacts with each other.

alexandr402 [8]3 years ago

6 0

Answer:

Physical

Explanation:

I'm not 100% sure on this, but since the chemical composition of the ingredients in the mixture did not change, it is not a chemical change. It is physical change because the texture and appearance change.

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A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The

hoa [83]

HA ⇄ H⁺ + A⁻
so:
$\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}$
and now:
$\frac{(x)(x)}{(0.150-x)}$ = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83

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3 years ago

Help plzzzzzzz ASAP!

Agata [3.3K]

I am going to say C. it has to do with the angles

8 0

3 years ago

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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

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3 years ago

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of $Fe_2O_3$ .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of $Fe_2O_3$

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of $Fe_2O_3$ required are, 22.1 grams.

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3 years ago

What is one molar volume of the gas ammonia (nh3) at stp?

Sedbober [7]

<span>The molar volume of ammonia at standard temperature and pressure (STP) is 22.4L. This is the standard molar volume of any gas at these conditions.</span>

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3 years ago

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