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notka56 [123]
4 years ago
9

Help Please earth science

Chemistry
1 answer:
alina1380 [7]4 years ago
4 0

Use the image to determine which of the following statements is true?

<span><span><span>1) The granite is younger than unit B.</span><span>2) Unit B and the granite are the same age.</span><span>3) The relative ages of teh granite and unit B cannot be determined from the information given.</span><span>4) Unit B is younger than the granite.</span></span></span>
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Amy found a volume of red liquid to be 55.1 ml, and have a mass of 8.199g. What is the density of her
NeX [460]

Answer: 0.15 g/mL

Explanation:

Density = mass/vol

3 0
3 years ago
MULTIPLE CHOICE<br> What is the molality (m) of 3.5 mol of sugar in 2.5 kg of water?
otez555 [7]

The molality of the solution is 1.4 moles per kilogram (molal).

The molality of the solution is equal to the number of moles of sugar (solute) divided by the number of kilograms of water (solvent). Hence, the molality is determined below:

m = \frac{3.5\,mol}{2.5\,kg}

m = 1.4\,\frac{mol}{kg}

The molality of the solution is 1.4 moles per kilogram (molal).

We kindly invite to check this question on molality: brainly.com/question/4580605

5 0
3 years ago
Help me!
AnnZ [28]

Answer:

a

Explanation:

a is the answer i too

k the test

4 0
3 years ago
Read 2 more answers
The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IB
mixas84 [53]

Answer:

\large \boxed{\text{0.0120 atm }}

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

      Kp = 280

p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} &  &  \\0 & & 0 & & p & & \\+x &   & +x  & & -2x &   &\\x &   & x} &   & 280 & & \\\end{array}

2. Calculate p(I₂)

\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.0120 atm }}$}}

Check:

\begin{array}{rcl}{\dfrac{0.200^{2}}{0.0120^{2}}}&=&280\\\\280& =& 280\\\end{array}

8 0
3 years ago
A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve
Helen [10]

Answer:

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

Explanation:

The differential equation is given as:

\frac{dC}{dt} = r- kC

\frac{dC}{r- kC} = dt

Taking integral of both sides; we have:

\int\limits  \frac{dC}{r- kC} = \int\limits dt

-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD

r-kC=e^{-kt-kD}

r-kC=e^{-kt}e^{-kD}

r-kC=Ae^{-kt}

kC=r-Ae^{-kt}

C=\frac{r}{k}-\frac{A}{k}e^{-kt}

C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}     ------- equation (1)

If C(0)= C_o ; we have:

C(0)= \frac{r}{k}-\frac{A}{k}e^0         (where; A is an integration constant)

C_o = \frac{r}{k}- \frac{A}{k}

C_o=\frac{r-A}{k}

kC_o=r-A

A=r-kC_o

Substituting A=r-kC_o into equation (1); we have;

C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}

3 0
4 years ago
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